Algebra 2

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use \( f(x) = x^2 - 4x \) where a function is not named.

1. Find \( f(3) \).
   Show the full solution

   \( 9 - 12 \). \( -3 \)

2. Find \( f(-1) \).
   Show the full solution

   \( 1 + 4 \). 5

3. Find \( f(0) \).
   Show the full solution

   0

4. Find the domain of \( \dfrac{1}{x - 5} \).
   Show the full solution

   Only the zero denominator excludes anything. All reals except 5

5. Find the domain of \( \sqrt{x + 2} \).
   Show the full solution

   \( x + 2 \ge 0 \). \( x \ge -2 \)

6. Find \( f(t - 2) \) and simplify.
   Show the full solution

   \( (t - 2)^2 - 4(t - 2) = t^2 - 4t + 4 - 4t + 8 = t^2 - 8t + 12 \). Check at \( t = 3 \): the formula gives \( 9 - 24 + 12 = -3 \), and \( f(1) = 1 - 4 = -3 \). Agrees. \( t^2 - 8t + 12 \)

7. Find the domain of \( \dfrac{x + 1}{x^2 - 9} \).
   Show the full solution

   Factor the denominator: \( (x - 3)(x + 3) \), which is zero at \( x = 3 \) and \( x = -3 \). Domain: all reals except 3 and \( -3 \), or \( (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \). Note that the numerator plays no part: a zero numerator gives an output of zero, which is perfectly allowed. All reals except \( \pm 3 \)

8. Find the domain of \( \dfrac{\sqrt{2x + 6}}{x - 1} \).
   Show the full solution

   Radicand nonnegative: \( 2x + 6 \ge 0 \), so \( x \ge -3 \). Denominator nonzero: \( x \ne 1 \). Both must hold: \( [-3, 1) \cup (1, \infty) \). Check a boundary: at \( x = -3 \) the expression is \( \dfrac{0}{-4} = 0 \), which is fine, so \( -3 \) is included. \( [-3, 1) \cup (1, \infty) \)

9. Explain why \( f(x) \) is not \( f \) times \( x \), and why the distinction matters.
   Show the full solution

   \( f \) is not a number. It is a rule, and the notation \( f(x) \) names the output that rule produces from the input \( x \). The parentheses are part of the naming, not a multiplication sign. Why it matters. Reading it as multiplication makes several later statements come out wrong. A student who believes \( f(x) \) is a product will "simplify" \( \dfrac{f(2x)}{f(x)} \) to 2, which is almost never true. Take \( f(x) = x^2 \): then \( f(2x) = 4x^2 \) and the ratio is 4, not 2. Take \( f(x) = x + 1 \): the ratio is \( \dfrac{2x + 1}{x + 1} \), which is not constant at all. The same misreading produces \( f(a + b) = f(a) + f(b) \), which is the linearity error this course names. It holds only for functions of the form \( f(x) = mx \), and for nothing else. The reliable habit. Read \( f(x) \) aloud as "f of x" rather than "f x". It costs nothing and it keeps the two ideas apart. \( f \) is a rule, not a factor; reading it as a product produces the linearity error

10. A function is defined by the table: \( f(1) = 4 \), \( f(2) = 7 \), \( f(3) = 4 \), \( f(4) = 10 \). State its domain and range, and say whether \( f \) has an inverse function.
    Show the full solution

    Domain. The set of inputs the table defines: \( \{1, 2, 3, 4\} \). Range. The set of outputs produced, each listed once: \( \{4, 7, 10\} \). The value 4 appears twice in the table but the range is a set, so it is written once. Is it a function? Yes. Each input appears exactly once with exactly one output. Two inputs sharing an output is entirely allowed; it is two outputs sharing an input that is forbidden. Does it have an inverse function? No. An inverse would have to send 4 back to its input, but 4 came from both 1 and 3. The inverse relation would pair the input 4 with two different outputs, which is not a function. The general statement. A function has an inverse function exactly when no output is repeated, which is the horizontal line test of lesson 1.7. This table fails it. What could be done. Restricting the domain to \( \{1, 2, 4\} \), or to \( \{2, 3, 4\} \), removes the repetition and produces a function that does have an inverse. That is the same move that makes \( \sqrt{x} \) the inverse of \( x^2 \). Domain \( \{1,2,3,4\} \), range \( \{4,7,10\} \), no inverse function because 4 is repeated

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. State the range of \( f(x) = x^2 \).
   Show the full solution

   \( y \ge 0 \)

2. State the domain of \( f(x) = \sqrt[3]{x} \).
   Show the full solution

   Odd roots accept negatives. All reals

3. State the range of \( f(x) = \left| x \right| \).
   Show the full solution

   \( y \ge 0 \)

4. Name the asymptotes of \( f(x) = \dfrac{1}{x} \).
   Show the full solution

   \( x = 0 \) and \( y = 0 \)

5. Which parent function has a corner rather than a smooth turn?
   Show the full solution

   Absolute value

6. Compare the graphs of \( x^2 \) and \( x^3 \) for \( x \) between 0 and 1.
   Show the full solution

   On \( 0 \lt x \lt 1 \), raising to a higher power makes a number smaller, since multiplying by a fraction reduces it. At \( x = 0.5 \): \( x^2 = 0.25 \) and \( x^3 = 0.125 \). So the cubic lies below the quadratic on that interval, which reverses outside it: at \( x = 2 \), \( x^2 = 4 \) and \( x^3 = 8 \). They cross at \( x = 0 \) and \( x = 1 \), where both equal 0 and 1 respectively. The cubic is below the quadratic on \( (0, 1) \) and above it beyond 1

7. State the domain and range of \( f(x) = 2^x \), and of \( f(x) = \log_2 x \).
   Show the full solution

   Exponential: domain all reals, range \( y \gt 0 \). No power of 2 is zero or negative, so the graph stays strictly above the \( x \)-axis. Logarithmic: domain \( x \gt 0 \), range all reals. Notice the swap: the exponential's domain is the logarithm's range and the exponential's range is the logarithm's domain. That is what being inverses means. Exponential: all reals, \( y \gt 0 \). Logarithmic: \( x \gt 0 \), all reals

8. Sketch-describe \( y = \sqrt{x} \) and \( y = x^2 \) on the same axes for \( x \ge 0 \), and say where they cross.
   Show the full solution

   Both start at the origin and rise. The square root rises quickly at first and then flattens; the parabola starts flat and then rises steeply. They cross where \( \sqrt{x} = x^2 \). Squaring both sides gives \( x = x^4 \), so \( x^4 - x = 0 \) and \( x(x^3 - 1) = 0 \), giving \( x = 0 \) or \( x = 1 \). Check both in the original: \( \sqrt{0} = 0 = 0^2 \) and \( \sqrt{1} = 1 = 1^2 \). Both genuine, neither extraneous. Between 0 and 1 the square root is above; beyond 1 the parabola is above. At \( x = 0.25 \): \( \sqrt{x} = 0.5 \) and \( x^2 = 0.0625 \). Confirmed. They cross at \( (0,0) \) and \( (1,1) \)

9. Explain why knowing these nine graphs makes the rest of the year easier.
   Show the full solution

   Every function in Algebra 2 is one of these nine with some combination of shifts, reflections and stretches applied, and lesson 1.3 shows that those transformations change a graph in entirely predictable ways. So graphing a new function is never a fresh problem. It is two questions: which parent is this, and what has been done to it. A student who can answer both can sketch \( y = -3\sqrt{x + 4} - 1 \) without plotting a single point. What that buys in each later unit. Unit 2's parabolas are the quadratic parent transformed. Unit 4's rational functions start from the reciprocal parent. Unit 5's radical functions are the square root and cube root parents. Units 6 and 7 are the exponential and logarithmic parents. Unit 9's sinusoids add one more parent and reuse the same four transformations. The deeper point. Organizing the course this way is a claim about mathematics, not a study tip: the transformations form a structure that acts the same way on every function, so learning it once covers every family. That is why this unit comes first rather than being tucked into whichever chapter happens to need it. Every later function is one of these nine transformed, so graphing reduces to identifying the parent and the transformation

10. For each of \( x^2 \), \( x^3 \), \( \sqrt{x} \) and \( \dfrac{1}{x} \), state whether it has an inverse function on its natural domain, and if not, what restriction fixes it.
    Show the full solution

    The test. A function has an inverse function exactly when no output is repeated, which on a graph is the horizontal line test: no horizontal line may meet the graph twice. \( x^2 \). Fails. The line \( y = 4 \) meets it at \( x = 2 \) and \( x = -2 \). Restricting the domain to \( x \ge 0 \) removes the left branch and makes it one-to-one; the inverse is then \( \sqrt{x} \). Restricting to \( x \le 0 \) works equally well and gives \( -\sqrt{x} \). \( x^3 \). Passes. The cubic is increasing everywhere, so no output repeats, and its inverse is \( \sqrt[3]{x} \) on all reals with no restriction needed. \( \sqrt{x} \). Passes on its natural domain \( x \ge 0 \), since it is increasing. Its inverse is \( x^2 \) restricted to \( x \ge 0 \), and that restriction on the inverse is the mirror image of the restriction that was needed on \( x^2 \) in the first place. \( \dfrac{1}{x} \). Passes. Although it has two branches, no output is repeated: the right branch produces only positive values and the left only negative. Its inverse is itself, since solving \( y = \dfrac{1}{x} \) for \( x \) gives \( x = \dfrac{1}{y} \). A function equal to its own inverse is called an involution, and its graph is symmetric about the line \( y = x \). The pattern. The even-powered and even-root functions need restrictions; the odd-powered and odd-root ones do not. It is the same sign-destroying property that separated \( \sqrt{x} \) from \( \sqrt[3]{x} \) in the worked example. \( x^3 \), \( \sqrt{x} \) and \( \frac{1}{x} \) have inverses as they stand; \( x^2 \) needs its domain restricted to \( x \ge 0 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Describe the transformation from \( x^2 \) to \( x^2 + 4 \).
   Show the full solution

   Up 4

2. Describe the transformation from \( x^2 \) to \( (x - 5)^2 \).
   Show the full solution

   Right 5

3. Describe the transformation from \( \left| x \right| \) to \( -\left| x \right| \).
   Show the full solution

   Reflection in the \( x \)-axis

4. Describe the transformation from \( \sqrt{x} \) to \( 4\sqrt{x} \).
   Show the full solution

   Vertical stretch by 4

5. Give the vertex of \( y = (x + 1)^2 - 7 \).
   Show the full solution

   \( (-1, -7) \)

6. Describe \( y = \dfrac{1}{x - 2} + 3 \) as a transformation, and give its asymptotes.
   Show the full solution

   From the reciprocal parent: right 2, up 3. The parent's asymptotes are \( x = 0 \) and \( y = 0 \). Moving right 2 carries the vertical asymptote to \( x = 2 \); moving up 3 carries the horizontal one to \( y = 3 \). Check: at \( x = 3 \), \( y = 1 + 3 = 4 \). At \( x = 102 \), \( y = 0.01 + 3 \), close to 3. Consistent with a horizontal asymptote at 3. Right 2 and up 3; asymptotes \( x = 2 \) and \( y = 3 \)

7. Describe \( y = \sqrt{3x + 12} \) as a transformation of \( \sqrt{x} \).
   Show the full solution

   Factor the inside first: \( \sqrt{3x + 12} = \sqrt{3(x + 4)} \). So \( b = 3 \) and \( h = -4 \): a horizontal compression by \( \dfrac{1}{3} \), then a shift 4 units left. Check the starting point: the radicand is zero at \( x = -4 \), and the graph starts there. Reading the shift as 12 would have put the start at \( x = -12 \), where the radicand is \( -24 \) and the function is undefined. Compression by \( \frac{1}{3} \), then left 4

8. Write the equation of \( y = x^3 \) reflected in the \( y \)-axis, then moved 1 right and 2 up. Simplify if possible.
   Show the full solution

   Reflection in the \( y \)-axis replaces \( x \) with \( -x \): \( y = (-x)^3 = -x^3 \). Right 1: replace \( x \) with \( x - 1 \), giving \( y = -(x - 1)^3 \). Up 2: \( y = -(x - 1)^3 + 2 \). A simplification worth noticing. Since the cubic is odd, \( -(x-1)^3 = (1 - x)^3 \), so the answer can also be written \( y = (1 - x)^3 + 2 \). Both are correct; the first shows the transformations more clearly. Check at \( x = 1 \): \( y = 0 + 2 = 2 \). The parent's inflection point at the origin has moved to \( (1, 2) \), as the shifts require. Check at \( x = 2 \): \( y = -(1) + 2 = 1 \), below the inflection point, which is right for a reflected cubic. \( y = -(x - 1)^3 + 2 \)

9. Explain why \( f(x) + 3 \) and \( f(x + 3) \) are different transformations.
   Show the full solution

   They act at different stages, and that is the whole difference. \( f(x) + 3 \) feeds \( x \) into the function, gets an answer, then adds 3 to that answer. Every output rises by 3, so the graph moves up 3. \( f(x + 3) \) adds 3 to the input before the function sees it. The function then behaves at \( x = -3 \) the way it used to behave at \( 0 \), so the graph moves left 3. A concrete comparison. Take \( f(x) = x^2 \). \( f(x) + 3 = x^2 + 3 \), with vertex \( (0, 3) \). \( f(x + 3) = (x + 3)^2 \), with vertex \( (-3, 0) \). Different graphs entirely: one moved vertically, one horizontally, in opposite apparent directions from their signs. The general rule this illustrates. Outside the function means the output, and acts as written. Inside the function means the input, and acts opposite. Everything in this lesson is that one distinction applied four times. One adds to the output and moves the graph up; the other adds to the input and moves it left

10. A graph of \( y = f(x) \) passes through \( (2, 5) \) and \( (-1, 0) \). Find the corresponding points on \( y = -2f(x - 1) + 4 \).
    Show the full solution

    Work out what happens to a general point. Suppose \( (p, q) \) is on \( y = f(x) \), so \( f(p) = q \). For the new function to use that value, its input must satisfy \( x - 1 = p \), so \( x = p + 1 \). The \( x \)-coordinate increases by 1. The output is then \( -2q + 4 \). So \( (p, q) \) maps to \( (p + 1,\; -2q + 4) \). Apply it to \( (2, 5) \). New \( x \): \( 2 + 1 = 3 \). New \( y \): \( -2(5) + 4 = -6 \). The point is \( (3, -6) \). Apply it to \( (-1, 0) \). New \( x \): \( -1 + 1 = 0 \). New \( y \): \( -2(0) + 4 = 4 \). The point is \( (0, 4) \). Check the order of operations on the output. The stretch and reflection apply before the shift, because the form is \( -2f(\cdot) + 4 \): multiply first, then add. Adding 4 first and then multiplying would give \( -2(5 + 4) = -18 \), which is wrong. Sanity check on the \( x \)-coordinate. The transformation is a shift right by 1, so both \( x \)-coordinates should increase by exactly 1, and they do. A note on what does not change. The horizontal shift does not touch the \( y \)-coordinate and the vertical work does not touch the \( x \)-coordinate. Keeping the two columns separate is the reliable way to do these. \( (3, -6) \) and \( (0, 4) \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Classify \( f(x) = x^2 \).
   Show the full solution

   Even

2. Classify \( f(x) = x^5 \).
   Show the full solution

   Odd

3. Classify \( f(x) = \left| x \right| \).
   Show the full solution

   \( \left| -x \right| = \left| x \right| \). Even

4. Classify \( f(x) = x + 2 \).
   Show the full solution

   \( f(-x) = -x + 2 \), which is neither \( f(x) \) nor \( -f(x) \). Neither

5. What symmetry does an even function's graph have?
   Show the full solution

   Symmetry about the \( y \)-axis

6. Classify \( f(x) = \dfrac{1}{x^3} \).
   Show the full solution

   \( f(-x) = \dfrac{1}{(-x)^3} = \dfrac{1}{-x^3} = -\dfrac{1}{x^3} = -f(x) \). Odd Check: \( f(2) = \dfrac{1}{8} \) and \( f(-2) = -\dfrac{1}{8} \). Opposite. Correct.

7. Classify \( f(x) = x^4 + x^2 - 6 \).
   Show the full solution

   \( f(-x) = x^4 + x^2 - 6 = f(x) \), since both powers are even and the constant is unchanged. Even The shortcut confirms it: every exponent present is even, counting \( -6 \) as \( -6x^0 \). Check: \( f(1) = 1 + 1 - 6 = -4 \) and \( f(-1) = -4 \). Equal.

8. Classify \( f(x) = \dfrac{x^2}{x^3 - x} \), and state its domain.
   Show the full solution

   Domain first. The denominator factors as \( x(x^2 - 1) = x(x-1)(x+1) \), zero at \( 0 \), \( 1 \) and \( -1 \). Domain: all reals except those three. Note the domain is symmetric about zero, which it must be for either classification to be possible. Test. \( f(-x) = \dfrac{(-x)^2}{(-x)^3 - (-x)} = \dfrac{x^2}{-x^3 + x} = \dfrac{x^2}{-(x^3 - x)} = -\dfrac{x^2}{x^3 - x} = -f(x) \). Odd Check at \( x = 2 \): \( f(2) = \dfrac{4}{8 - 2} = \dfrac{2}{3} \), and \( f(-2) = \dfrac{4}{-8 + 2} = -\dfrac{2}{3} \). Opposite. Correct. Note on the shortcut. The all-even or all-odd exponent rule applies to polynomials, not to quotients. Here an even numerator over an odd denominator produced an odd function, which the rule would not have predicted.

9. Explain why a function whose domain is \( x \ge 0 \) cannot be even or odd, unless its domain is just \( \{0\} \).
   Show the full solution

   Both tests require evaluating \( f(-x) \). For the test to mean anything, \( -x \) must be in the domain whenever \( x \) is. A domain of \( x \ge 0 \) fails that at once: if \( x = 2 \) is in the domain then \( -2 \) must be too, and it is not. So \( f(-2) \) does not exist and the equation \( f(-2) = f(2) \) cannot hold or fail; it is not a statement about anything. The general requirement. A function can be even or odd only if its domain is symmetric about zero. That is a precondition, checked before any algebra. Why \( \{0\} \) is the exception. The set \( \{0\} \) is symmetric about zero, since \( -0 = 0 \). A function defined only there is even if \( f(0) \) is anything, and odd only if \( f(0) = 0 \), since oddness would demand \( f(0) = -f(0) \). A worked case. \( f(x) = \sqrt{x} \) has domain \( x \ge 0 \) and is therefore neither, for this reason rather than for any algebraic one. By contrast \( f(x) = \sqrt[3]{x} \) has all reals as its domain and is genuinely odd. The practical order. Check the domain for symmetry first. If it is not symmetric, the answer is "neither" and no computation is needed. The tests require \( -x \) to be in the domain whenever \( x \) is, and a domain of \( x \ge 0 \) is not symmetric about zero

10. Prove that the product of two odd functions is even, and that the product of an even and an odd function is odd.
    Show the full solution

    Set up. Let \( f \) and \( g \) be functions on a domain symmetric about zero, and let \( p(x) = f(x)g(x) \). First claim: odd times odd is even. Suppose \( f \) and \( g \) are both odd, so \( f(-x) = -f(x) \) and \( g(-x) = -g(x) \). Then \[ p(-x) = f(-x)\,g(-x) = \big(-f(x)\big)\big(-g(x)\big) = f(x)g(x) = p(x) \] The two negatives multiply to a positive, so \( p \) is even. Check with an example. \( f(x) = x \) and \( g(x) = x^3 \) are both odd, and their product is \( x^4 \), which is even. Confirmed. Second claim: even times odd is odd. Suppose \( f \) is even and \( g \) is odd, so \( f(-x) = f(x) \) and \( g(-x) = -g(x) \). Then \[ p(-x) = f(-x)\,g(-x) = f(x)\big(-g(x)\big) = -f(x)g(x) = -p(x) \] One negative survives, so \( p \) is odd. Check with an example. \( f(x) = x^2 \) is even and \( g(x) = x \) is odd, and their product is \( x^3 \), which is odd. Confirmed. The third case, for completeness. Even times even is even, by the same argument with no negatives at all. So the three rules together behave exactly like multiplying signs, or like adding exponents in \( x^m \cdot x^n = x^{m+n} \) and asking whether the sum is even or odd. Why the analogy is not a coincidence. The names come from the power functions, where evenness and oddness are literally the parity of the exponent. Since multiplying powers adds exponents, and parity adds the way these rules multiply, the general result had to come out matching. A caution. The corresponding statements for sums are weaker: even plus even is even and odd plus odd is odd, but even plus odd is usually neither. \( x^2 + x \) from the worked example is exactly that case. Proved by substituting the definitions and tracking the signs

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( \left| x \right| = 6 \).
   Show the full solution

   \( x = 6 \) or \( x = -6 \)

2. Solve \( \left| x \right| = -2 \).
   Show the full solution

   An absolute value is never negative. No solution

3. Solve \( \left| x \right| \lt 5 \).
   Show the full solution

   \( -5 \lt x \lt 5 \)

4. Solve \( \left| x \right| \gt 2 \).
   Show the full solution

   Two rays. \( x \lt -2 \) or \( x \gt 2 \)

5. For \( f(x) = \begin{cases} x + 3 & x \le 0 \\ 5 & x \gt 0 \end{cases} \), find \( f(-2) \).
   Show the full solution

   \( -2 \le 0 \), so use the first rule. 1

6. Solve \( \left| 4x - 1 \right| = 11 \).
   Show the full solution

   \( 4x - 1 = 11 \) gives \( x = 3 \). \( 4x - 1 = -11 \) gives \( 4x = -10 \) and \( x = -\dfrac{5}{2} \). Check: \( \left| 12 - 1 \right| = 11 \). Correct. \( \left| -10 - 1 \right| = 11 \). Correct. \( x = 3 \) or \( x = -\frac{5}{2} \)

7. Solve \( \left| 2x + 7 \right| \le 3 \).
   Show the full solution

   \( -3 \le 2x + 7 \le 3 \). Subtract 7: \( -10 \le 2x \le -4 \). Divide by 2: \( -5 \le x \le -2 \). Check an endpoint: at \( x = -5 \), \( \left| -3 \right| = 3 \le 3 \). Included, correctly. Check the middle: at \( x = -3.5 \), \( \left| 0 \right| = 0 \le 3 \). Correct. \( [-5, -2] \)

8. Write \( f(x) = \left| x - 3 \right| \) as a piecewise function, and give its vertex.
   Show the full solution

   The absolute value returns its input when the input is nonnegative and the opposite otherwise, so split at the point where \( x - 3 = 0 \), that is \( x = 3 \). \[ f(x) = \begin{cases} x - 3 & x \ge 3 \\ 3 - x & x \lt 3 \end{cases} \] Note the second piece: \( -(x - 3) = 3 - x \), not \( -x - 3 \). Distributing the negative across both terms is where this is usually lost. Vertex. The corner is where the two pieces meet, at \( x = 3 \), with \( f(3) = 0 \). Vertex \( (3, 0) \). Check: \( f(5) = 2 \) from the first piece, and \( \left| 5 - 3 \right| = 2 \). Correct. \( f(1) = 3 - 1 = 2 \) from the second, and \( \left| 1 - 3 \right| = 2 \). Correct. Piecewise as above, vertex \( (3, 0) \)

9. Explain why \( \left| A \right| \lt b \) gives one interval but \( \left| A \right| \gt b \) gives two.
   Show the full solution

   Read \( \left| A \right| \) as the distance from \( A \) to zero on the number line. The "less than" case. Asking for points within \( b \) of zero describes a single stretch centered on zero, from \( -b \) to \( b \). That is one interval, and it can be written as the chain \( -b \lt A \lt b \) because every point in it satisfies both halves at once. The "greater than" case. Asking for points more than \( b \) from zero describes everything outside that stretch, and the outside of an interval on a line is two pieces: one running left from \( -b \) and one running right from \( b \). Nothing lies in both, so they are joined by "or" rather than "and". Why the chain notation fails there. Writing \( -b \gt A \gt b \) reads as \( A \) being simultaneously less than \( -b \) and greater than \( b \), which no number is when \( b \) is positive. The notation asserts an impossibility, so it can never be right. The check that catches it. Test a value. For \( \left| x \right| \gt 3 \), the number 5 should work: \( \left| 5 \right| = 5 \gt 3 \). Substituting into \( -3 \gt x \gt 3 \) gives \( -3 \gt 5 \), which is false, so the notation has excluded a genuine solution. Less than means inside a single centered interval; greater than means outside it, and the outside of an interval is two pieces

10. A machine part must be 12.0 cm long with a tolerance of 0.4 cm. Write this as an absolute value inequality, solve it, and then write the acceptable range as a piecewise description of a "reject" function.
    Show the full solution

    Write the inequality. "Within 0.4 of 12.0" is a statement about distance, so it is an absolute value: \( \left| L - 12.0 \right| \le 0.4 \), where \( L \) is the measured length in centimeters. Solve it. \( -0.4 \le L - 12.0 \le 0.4 \). Add 12.0 throughout: \( 11.6 \le L \le 12.0 + 0.4 = 12.4 \). Acceptable range: \( [11.6,\; 12.4] \). Check the endpoints. At \( L = 11.6 \), \( \left| -0.4 \right| = 0.4 \le 0.4 \). Accepted, correctly, since the tolerance is inclusive. At \( L = 12.5 \), \( \left| 0.5 \right| = 0.5 \), which exceeds 0.4. Rejected. Correct. The reject function. Define \( R(L) \) as 1 for a rejected part and 0 for an accepted one: \[ R(L) = \begin{cases} 1 & L \lt 11.6 \\ 0 & 11.6 \le L \le 12.4 \\ 1 & L \gt 12.4 \end{cases} \] This is a genuine piecewise function with three pieces and two jump discontinuities, at 11.6 and at 12.4. Why the boundary convention matters here. The inequalities on the middle piece are inclusive, which is a real decision with a cost: a part measuring exactly 12.4 is accepted. Writing the middle piece with strict inequalities would reject it. In manufacturing that convention is written into the specification precisely because measurement noise makes borderline parts common. A note on the model. The function treats measurement as exact. A real inspection has its own uncertainty, so a part measured at 12.41 might truly be 12.39. That is why tolerances in practice are set tighter than the true requirement, and it is the kind of assumption a modeling answer should state rather than hide. \( \left| L - 12.0 \right| \le 0.4 \), giving \( 11.6 \le L \le 12.4 \), with the three-piece reject function above

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use \( f(x) = 3x + 1 \) and \( g(x) = x^2 \) unless told otherwise.

1. Find \( f(g(2)) \).
   Show the full solution

   \( g(2) = 4 \), then \( f(4) = 13 \). 13

2. Find \( g(f(2)) \).
   Show the full solution

   \( f(2) = 7 \), then \( g(7) = 49 \). 49

3. Find \( (f \circ g)(x) \).
   Show the full solution

   \( f(x^2) = 3x^2 + 1 \). \( 3x^2 + 1 \)

4. Find \( (g \circ f)(x) \).
   Show the full solution

   \( g(3x + 1) = (3x + 1)^2 \). \( (3x+1)^2 = 9x^2 + 6x + 1 \)

5. Which function is applied first in \( (f \circ g)(x) \)?
   Show the full solution

   \( g \)

6. With \( f(x) = \dfrac{1}{x} \) and \( g(x) = x + 4 \), find \( (f \circ g)(x) \) and its domain.
   Show the full solution

   \( f(g(x)) = f(x + 4) = \dfrac{1}{x + 4} \). Domain: \( g \) accepts everything, and \( f \) rejects an input of zero, so \( x + 4 \ne 0 \) and \( x \ne -4 \). Domain: all reals except \( -4 \). Check: at \( x = -4 \) the expression is \( \dfrac{1}{0} \), undefined. Correctly excluded. \( \frac{1}{x+4} \), all reals except \( -4 \)

7. With \( f(x) = \sqrt{x} \) and \( g(x) = x^2 \), find \( (f \circ g)(x) \) and simplify carefully.
   Show the full solution

   \( f(g(x)) = \sqrt{x^2} \). The simplification is not \( x \). The radical denotes the principal root, which is never negative, so \( \sqrt{x^2} = \left| x \right| \). Check with a negative input: at \( x = -3 \), \( \sqrt{9} = 3 \), not \( -3 \). So the answer must be \( \left| x \right| \). Domain: \( g \) accepts every real, and its output \( x^2 \) is never negative, so \( f \) accepts it. Domain: all reals. This is the even-root absolute value that appears throughout unit 5. \( \left| x \right| \), domain all reals

8. With \( f(x) = 2x - 6 \) and \( g(x) = \dfrac{x + 6}{2} \), find both compositions and say what you notice.
   Show the full solution

   \( (f \circ g)(x) = f\left( \dfrac{x+6}{2} \right) = 2 \cdot \dfrac{x+6}{2} - 6 = (x + 6) - 6 = x \). \( (g \circ f)(x) = g(2x - 6) = \dfrac{(2x - 6) + 6}{2} = \dfrac{2x}{2} = x \). Both compositions give \( x \), the identity function. What that means. Each function exactly undoes the other, so they are inverses. That is the definition lesson 1.7 uses, and it is the one case where the order of composition genuinely does not matter. Check numerically: \( f(5) = 4 \) and \( g(4) = 5 \). Round trip confirmed. Both equal \( x \); the functions are inverses

9. Explain why the domain of a composition must be found before simplifying.
   Show the full solution

   Simplification can cancel the very expression that caused a restriction, leaving a formula that looks unrestricted while the function still is. A concrete case. Let \( f(x) = x^2 \) and \( g(x) = \dfrac{1}{x} \). Then \( (f \circ g)(x) = \left( \dfrac{1}{x} \right)^2 = \dfrac{1}{x^2} \), which still shows the restriction. But take \( f(x) = \dfrac{1}{x} \) and \( g(x) = \dfrac{1}{x} \): the composition simplifies to \( x \), and the final formula shows no restriction at all. Yet \( x = 0 \) is not in the domain, because the inner function cannot accept it. The composition is the identity on every input except zero, where it is undefined. Why this is not a technicality. Two functions are equal only when they have the same domain and the same rule. The composition above and the identity function \( y = x \) are different functions, and a graph of the composition has a hole at the origin. The reliable procedure. First require that \( x \) is in the domain of the inner function. Then require that the inner output is in the domain of the outer function. Only then simplify, and never let the simplified form revise the domain. Canceling can hide a restriction that the inner function still imposes, so the domain is read from the composition rather than from the simplified formula

10. A store applies a 20 percent discount and then adds 8 percent sales tax. Write both operations as functions, compose them in both orders, and say whether the order matters.
    Show the full solution

    Write the functions. Let \( x \) be the marked price in dollars. Discount: paying 80 percent of the price, so \( d(x) = 0.80x \). Tax: adding 8 percent, so \( t(x) = 1.08x \). Discount first, then tax. \( (t \circ d)(x) = t(0.80x) = 1.08(0.80x) = 0.864x \). Tax first, then discount. \( (d \circ t)(x) = d(1.08x) = 0.80(1.08x) = 0.864x \). The order does not matter here. Both give \( 0.864x \), that is 86.4 percent of the marked price. Check with a number. Take \( x = 100 \). Discount first: \( 100 \to 80 \to 86.40 \). Tax first: \( 100 \to 108 \to 86.40 \). Identical. Confirmed. Why the order does not matter in this case. Both operations are multiplications by a constant, and multiplication is commutative: \( 1.08 \times 0.80 = 0.80 \times 1.08 \). Composition of two such functions is just multiplication of their factors. When the order would matter. Change the discount to a flat $20 off, so \( d(x) = x - 20 \). Then \( (t \circ d)(100) = t(80) = 86.40 \), while \( (d \circ t)(100) = d(108) = 88 \). Now the order costs the customer $1.60, because one operation is multiplicative and the other additive, and those do not commute. The general lesson. Composition commutes only in special circumstances, and "both operations scale by a constant" is one of the few. Whenever a problem mixes a percentage with a flat amount, the order has to be stated, which is why real tax law specifies it explicitly. Both orders give \( 0.864x \); the order does not matter because both are scalings, but it would if either were a flat amount

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the inverse of \( f(x) = x + 7 \).
   Show the full solution

   \( f^{-1}(x) = x - 7 \)

2. Find the inverse of \( f(x) = 5x \).
   Show the full solution

   \( f^{-1}(x) = \frac{x}{5} \)

3. Find the inverse of \( f(x) = 2x - 8 \).
   Show the full solution

   \( x = 2y - 8 \), so \( y = \dfrac{x + 8}{2} \). \( f^{-1}(x) = \frac{x+8}{2} \)

4. Does \( f(x) = x^4 \) have an inverse function on all reals?
   Show the full solution

   An even power repeats outputs. No

5. What line are \( f \) and \( f^{-1} \) reflections in?
   Show the full solution

   \( y = x \)

6. Find the inverse of \( f(x) = \dfrac{x}{x - 1} \).
   Show the full solution

   Swap: \( x = \dfrac{y}{y - 1} \). Clear: \( x(y - 1) = y \), so \( xy - x = y \). Collect: \( xy - y = x \), so \( y(x - 1) = x \) and \( y = \dfrac{x}{x - 1} \). The function is its own inverse. Check: \( f(2) = \dfrac{2}{1} = 2 \), and \( f(3) = \dfrac{3}{2} \), then \( f\left( \dfrac{3}{2} \right) = \dfrac{3/2}{1/2} = 3 \). Round trip closed. \( f^{-1}(x) = \frac{x}{x-1} \), the same function

7. Find the inverse of \( f(x) = \sqrt{x - 4} \) and state its domain.
   Show the full solution

   Swap: \( x = \sqrt{y - 4} \). Square both sides: \( x^2 = y - 4 \), so \( y = x^2 + 4 \). The domain restriction is essential. The original function outputs only nonnegative values, so its range is \( y \ge 0 \), which becomes the domain of the inverse. \( f^{-1}(x) = x^2 + 4 \) for \( x \ge 0 \). Without that restriction, \( x^2 + 4 \) would not be an inverse: at \( x = -3 \) it gives 13, but \( f(13) = \sqrt{9} = 3 \), not \( -3 \). Check with a valid value: \( f(20) = \sqrt{16} = 4 \), and \( f^{-1}(4) = 16 + 4 = 20 \). Correct. \( f^{-1}(x) = x^2 + 4 \), domain \( x \ge 0 \)

8. Verify that \( f(x) = \dfrac{x - 5}{3} \) and \( g(x) = 3x + 5 \) are inverses.
   Show the full solution

   Both compositions must give the identity. \( f(g(x)) = \dfrac{(3x + 5) - 5}{3} = \dfrac{3x}{3} = x \). Correct. \( g(f(x)) = 3 \cdot \dfrac{x - 5}{3} + 5 = (x - 5) + 5 = x \). Correct. Why both are needed. One composition alone is not enough in general. A function can undo another in one direction only, when domains are restricted, so the definition requires both and a complete verification shows both. Numerical confirmation: \( g(2) = 11 \) and \( f(11) = \dfrac{6}{3} = 2 \). Verified: both compositions equal \( x \)

9. Explain why a function must be one-to-one to have an inverse function.
   Show the full solution

   An inverse must send each output back to the input it came from. If two different inputs produced the same output, the inverse would have to send that one output back to two different values, and a function is not permitted to do that. The definition it violates. A function assigns exactly one output to each input. The inverse relation would assign two, so it is a relation but not a function. A concrete case. For \( f(x) = x^2 \), both 4 and \( -4 \) map to 16. Asking for \( f^{-1}(16) \) has no single right answer, and choosing 4 by convention is what defines \( \sqrt{\cdot} \) rather than what the inverse would naturally be. Where this recurs this year. The same problem forces every restriction of the year: \( \sqrt{x} \) exists as a function only because \( x^2 \) was restricted to \( x \ge 0 \); the logarithm works without restriction because the exponential is already one-to-one; and the inverse sine, cosine and tangent of unit 10 need narrow restricted ranges because the trigonometric functions repeat their outputs infinitely often. The pattern worth extracting. Whenever a course introduces an inverse with an odd-looking restriction attached, the restriction is not arbitrary. It is the minimum needed to make the original function one-to-one, and it was chosen once, by convention, so that the inverse has a single agreed value. Two inputs sharing an output would force the inverse to assign two outputs to one input, which no function may do

10. A temperature conversion is \( F(c) = \dfrac{9}{5}c + 32 \). Find its inverse, interpret both functions, and find the temperature at which the two scales agree.
    Show the full solution

    Find the inverse. Write \( f = \dfrac{9}{5}c + 32 \) and solve for \( c \): \( f - 32 = \dfrac{9}{5}c \), so \( c = \dfrac{5}{9}(f - 32) \). As a function of its input, \( F^{-1}(f) = \dfrac{5}{9}(f - 32) \). Interpret. \( F \) converts Celsius to Fahrenheit; \( F^{-1} \) converts Fahrenheit back to Celsius. The inverse undoes the conversion, which is exactly what the word means here. Verify both ways. \( F(100) = \dfrac{9}{5}(100) + 32 = 180 + 32 = 212 \), the boiling point. Correct. \( F^{-1}(212) = \dfrac{5}{9}(180) = 100 \). Round trip closed. \( F(0) = 32 \) and \( F^{-1}(32) = 0 \), the freezing point. Correct. Find where the scales agree. The two scales give the same number when \( F(c) = c \): \( \dfrac{9}{5}c + 32 = c \). Multiply through by 5: \( 9c + 160 = 5c \). So \( 4c = -160 \) and \( c = -40 \). Check. \( F(-40) = \dfrac{9}{5}(-40) + 32 = -72 + 32 = -40 \). Correct. Minus forty degrees is the same temperature on both scales, which is why weather reports at that temperature omit the unit without ambiguity. The graphical meaning. Solving \( F(c) = c \) finds where the graph of \( F \) meets the line \( y = x \). Since \( F \) and \( F^{-1} \) are reflections in that line, any point where a function meets \( y = x \) is a point the function and its inverse share. So \( (-40, -40) \) lies on both graphs, and it is their only intersection here because both are lines with different slopes. A note on the model. The conversion is exact by definition rather than approximate, since both scales are defined by fixed reference points. That is unusual: most formulas in an applied problem are models with error, and this one is not. \( F^{-1}(f) = \frac{5}{9}(f - 32) \); the scales agree at \( -40^\circ \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Give the vertex of \( y = (x - 4)^2 + 3 \).
   Show the full solution

   \( (4, 3) \)

2. Give the vertex of \( y = (x + 2)^2 - 7 \).
   Show the full solution

   \( x + 2 \) means \( h = -2 \). \( (-2, -7) \)

3. Give the roots of \( y = (x - 3)(x + 5) \).
   Show the full solution

   3 and \( -5 \)

4. Give the \( y \)-intercept of \( y = 3x^2 - x + 7 \).
   Show the full solution

   7

5. Does \( y = -2x^2 + 5 \) open up or down?
   Show the full solution

   Down

6. Find the axis of symmetry of \( y = x^2 - 8x + 3 \) two ways.
   Show the full solution

   By formula: \( x = -\dfrac{b}{2a} = -\dfrac{-8}{2} = 4 \). By completing the square: \( x^2 - 8x + 16 - 16 + 3 = (x - 4)^2 - 13 \), so \( h = 4 \). Both give \( x = 4 \), and the vertex is \( (4, -13) \). Check: \( f(4) = 16 - 32 + 3 = -13 \). Correct. \( x = 4 \)

7. Write \( y = (x - 2)(x + 6) \) in standard form and give its vertex.
   Show the full solution

   Expand: \( x^2 + 6x - 2x - 12 = x^2 + 4x - 12 \). Axis: the midpoint of the roots 2 and \( -6 \) is \( \dfrac{2 + (-6)}{2} = -2 \). Evaluate there: \( (-2)^2 + 4(-2) - 12 = 4 - 8 - 12 = -16 \). Vertex \( (-2, -16) \). Check with the formula: \( -\dfrac{4}{2} = -2 \). Agrees. \( y = x^2 + 4x - 12 \), vertex \( (-2, -16) \)

8. A parabola has roots at \( -1 \) and 7 and passes through \( (0, -14) \). Find its equation in all three forms.
   Show the full solution

   Start from factored form, since the roots are given: \( y = a(x + 1)(x - 7) \). Find \( a \) using the extra point. At \( x = 0 \), \( y = -14 \): \( -14 = a(1)(-7) = -7a \), so \( a = 2 \). Factored: \( y = 2(x + 1)(x - 7) \). Standard: \( 2(x^2 - 6x - 7) = 2x^2 - 12x - 14 \). Vertex: the axis is the midpoint of \( -1 \) and 7, which is 3. \( f(3) = 2(9) - 36 - 14 = 18 - 50 = -32 \). Vertex form: \( y = 2(x - 3)^2 - 32 \). Check all three at \( x = 0 \): factored \( 2(1)(-7) = -14 \); standard \( -14 \); vertex \( 2(9) - 32 = -14 \). All agree. \( 2(x+1)(x-7) = 2x^2 - 12x - 14 = 2(x-3)^2 - 32 \)

9. Explain why a quadratic with no real roots cannot be written in factored form over the real numbers.
   Show the full solution

   Factored form is \( a(x - r_1)(x - r_2) \), and substituting \( x = r_1 \) makes the first factor zero, so the whole expression is zero. That means \( r_1 \) is an \( x \)-intercept, and likewise \( r_2 \). So a factored form over the reals exists only when real \( x \)-intercepts exist. A parabola sitting entirely above the \( x \)-axis has none, so no such factorization is available. A concrete case. \( y = x^2 + 1 \) has minimum value 1 and never reaches zero. Any real factorization \( (x - r_1)(x - r_2) \) would produce a root, so none exists. What lesson 2.6 adds. The factorization does exist over the complex numbers: \( x^2 + 1 = (x - i)(x + i) \), and expanding confirms it, since \( x^2 - i^2 = x^2 + 1 \). The roots are \( \pm i \), which are not points on the real graph but are genuine solutions of the equation. The distinction worth keeping. "No solution" and "no real solution" are different statements. Algebra 1 said the first; Algebra 2 corrects it to the second, and this unit supplies the numbers that make the correction possible. Factored form would exhibit real roots, and a parabola with no \( x \)-intercepts has none; it factors over the complex numbers instead

10. A parabola has vertex \( (2, -9) \) and passes through \( (5, 0) \). Find all three forms and both roots.
    Show the full solution

    Start from vertex form, since the vertex is given: \( y = a(x - 2)^2 - 9 \). Find \( a \) with the extra point. At \( x = 5 \), \( y = 0 \): \( 0 = a(3)^2 - 9 = 9a - 9 \), so \( a = 1 \). Vertex form: \( y = (x - 2)^2 - 9 \). Standard form. Expand: \( x^2 - 4x + 4 - 9 = x^2 - 4x - 5 \). Factored form. Two numbers multiplying to \( -5 \) and adding to \( -4 \) are \( -5 \) and \( 1 \): \( y = (x - 5)(x + 1) \). Roots: \( x = 5 \) and \( x = -1 \). Check the roots against the vertex. Their midpoint is \( \dfrac{5 + (-1)}{2} = 2 \), which is the given \( h \). Consistent. Check all three forms at \( x = 0 \). Vertex: \( (0-2)^2 - 9 = 4 - 9 = -5 \). Standard: \( -5 \). Factored: \( (-5)(1) = -5 \). All agree. A second route worth seeing. Because the vertex is a minimum 9 units below the axis and \( a = 1 \), the roots sit \( \sqrt{9} = 3 \) units either side of \( x = 2 \), giving \( -1 \) and 5 directly. In general, solving \( a(x-h)^2 + k = 0 \) gives \( x = h \pm \sqrt{-k/a} \), which is why a vertex below the axis on an upward parabola always produces two real roots symmetric about the axis. \( (x-2)^2 - 9 = x^2 - 4x - 5 = (x-5)(x+1) \), roots 5 and \( -1 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What constant completes the square for \( x^2 + 10x \)?
   Show the full solution

   Half of 10 is 5, squared. 25

2. What constant completes the square for \( x^2 - 14x \)?
   Show the full solution

   Half of \( -14 \) is \( -7 \), squared. 49

3. Write \( x^2 + 6x + 9 \) as a square.
   Show the full solution

   \( (x + 3)^2 \)

4. Solve \( (x - 2)^2 = 25 \).
   Show the full solution

   \( x - 2 = \pm 5 \). \( x = 7 \) or \( x = -3 \)

5. Write \( x^2 - 4x \) in the form \( (x - p)^2 + q \).
   Show the full solution

   \( x^2 - 4x + 4 - 4 \). \( (x - 2)^2 - 4 \)

6. Write \( x^2 + 5x + 1 \) in vertex form.
   Show the full solution

   Half of 5 is \( \dfrac{5}{2} \), squared is \( \dfrac{25}{4} \). \( x^2 + 5x + \dfrac{25}{4} - \dfrac{25}{4} + 1 = \left( x + \dfrac{5}{2} \right)^2 - \dfrac{21}{4} \). Check at \( x = 0 \): the original gives 1, and \( \dfrac{25}{4} - \dfrac{21}{4} = \dfrac{4}{4} = 1 \). Correct. An odd linear coefficient produces fractions, and that is normal rather than a sign of error. \( \left( x + \frac{5}{2} \right)^2 - \frac{21}{4} \)

7. Solve \( x^2 + 10x + 7 = 0 \) by completing the square.
   Show the full solution

   \( x^2 + 10x = -7 \). Half of 10 is 5, squared is 25. Add to both sides: \( x^2 + 10x + 25 = 18 \), so \( (x + 5)^2 = 18 \). \( x + 5 = \pm\sqrt{18} = \pm 3\sqrt{2} \). \( x = -5 \pm 3\sqrt{2} \). Check exactly with \( -5 + 3\sqrt{2} \): \( (-5 + 3\sqrt{2})^2 = 25 - 30\sqrt{2} + 18 = 43 - 30\sqrt{2} \). \( 10(-5 + 3\sqrt{2}) = -50 + 30\sqrt{2} \). Sum with the 7: \( 43 - 30\sqrt{2} - 50 + 30\sqrt{2} + 7 = 0 \). Correct. \( x = -5 \pm 3\sqrt{2} \)

8. Write \( -2x^2 + 8x - 1 \) in vertex form and give the maximum value.
   Show the full solution

   Factor out \( -2 \) from the first two terms: \( -2(x^2 - 4x) - 1 \). Half of \( -4 \) is \( -2 \), squared is 4: \( -2(x^2 - 4x + 4 - 4) - 1 = -2\big[(x - 2)^2 - 4\big] - 1 \). Distribute: \( -2(x - 2)^2 + 8 - 1 = -2(x - 2)^2 + 7 \). Watch the sign. The \( -4 \) inside became \( +8 \) outside, because \( -2 \times -4 = +8 \). Getting this sign wrong is the most common error with a negative leading coefficient. Vertex \( (2, 7) \), and since \( a = -2 \) is negative the parabola opens down, so 7 is a maximum. Check at \( x = 0 \): the original gives \( -1 \), and \( -2(4) + 7 = -1 \). Correct. Check at \( x = 2 \): \( -8 + 16 - 1 = 7 \). Correct. \( -2(x-2)^2 + 7 \), maximum value 7

9. Explain why completing the square works on every quadratic, while factoring does not.
   Show the full solution

   Completing the square is a construction, not a search. Given \( x^2 + bx \), the constant \( \left( \dfrac{b}{2} \right)^2 \) always exists as a real number, so the perfect square \( \left( x + \dfrac{b}{2} \right)^2 \) can always be formed. Nothing about the coefficients can prevent it. Factoring over the integers is a search for two integers with a given product and sum, and no such pair need exist. For \( x^2 - 6x + 2 \) from the worked example, no integers multiply to 2 and add to \( -6 \), so integer factoring fails while completing the square succeeds and yields \( 3 \pm \sqrt{7} \). The consequence for the course. Because the method never fails, it can be applied to the general quadratic \( ax^2 + bx + c = 0 \) with letters in place of numbers. Doing so produces the quadratic formula, which is lesson 2.3. A method that worked only sometimes could not have been used that way. The one thing it does not guarantee. Completing the square always produces \( (x + p)^2 = q \), but if \( q \) is negative there is no real square root and the solutions are complex. The method still works; it simply reports honestly that the roots are not real, which lessons 2.5 and 2.6 then take seriously rather than treating as a dead end. The completing constant always exists, so the method never fails; factoring requires a pair of integers that may not exist

10. Find the center and radius of the circle \( x^2 + y^2 - 10x + 4y + 13 = 0 \) by completing the square twice, and explain the connection to this lesson.
    Show the full solution

    Group the variables. \( (x^2 - 10x) + (y^2 + 4y) = -13 \). Complete the square in \( x \). Half of \( -10 \) is \( -5 \), squared is 25. Complete the square in \( y \). Half of 4 is 2, squared is 4. Add both to both sides: \( (x^2 - 10x + 25) + (y^2 + 4y + 4) = -13 + 25 + 4 = 16 \). \( (x - 5)^2 + (y + 2)^2 = 16 \). Read off. Center \( (5, -2) \), radius \( \sqrt{16} = 4 \). Check with a point. Moving 4 right of the center gives \( (9, -2) \). Substituting into the original: \( 81 + 4 - 90 - 8 + 13 = 0 \). Correct. Watch the two traps. The signs flip when reading the center, so \( (y + 2)^2 \) gives \( -2 \). And the right side is \( r^2 \), so the radius is 4, not 16. The connection. This is the identical algebraic move done twice, once per variable. The circle equation is the distance formula squared, and completing the square undoes the expansion that hid the center. Nothing new was learned here beyond applying a quadratic technique to a two-variable equation. Why that matters. Completing the square is not a quadratic-equation trick. It is the general method for converting any squared expression from expanded to centered form, and it reappears with conic sections in Precalculus and with integration techniques in calculus. Learning it as a construction rather than a recipe is what makes those later uses recognizable. Center \( (5, -2) \), radius 4

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Give exact answers.

1. Identify \( a \), \( b \) and \( c \) in \( x^2 - 7x + 2 = 0 \).
   Show the full solution

   \( a = 1 \), \( b = -7 \), \( c = 2 \)

2. Compute the discriminant of \( x^2 + 4x + 3 = 0 \).
   Show the full solution

   \( 16 - 12 \). 4

3. Solve \( x^2 - 5x + 6 = 0 \).
   Show the full solution

   Discriminant \( 25 - 24 = 1 \), so \( x = \dfrac{5 \pm 1}{2} \). \( x = 3 \) or \( x = 2 \)

4. Solve \( x^2 - 2x - 1 = 0 \).
   Show the full solution

   Discriminant \( 4 + 4 = 8 \), so \( x = \dfrac{2 \pm 2\sqrt{2}}{2} \). \( x = 1 \pm \sqrt{2} \)

5. What must be true before the formula is applied?
   Show the full solution

   The equation must be in standard form with zero on one side

6. Solve \( 2x^2 - 7x + 3 = 0 \).
   Show the full solution

   \( a = 2 \), \( b = -7 \), \( c = 3 \). Discriminant: \( 49 - 24 = 25 \), a perfect square. \( x = \dfrac{7 \pm 5}{4} \), giving \( x = 3 \) or \( x = \dfrac{1}{2} \). Check by sum and product: sum should be \( \dfrac{7}{2} \), and \( 3 + \dfrac{1}{2} = \dfrac{7}{2} \). Product should be \( \dfrac{3}{2} \), and \( 3 \times \dfrac{1}{2} = \dfrac{3}{2} \). Both correct. \( x = 3 \) or \( x = \frac{1}{2} \)

7. Solve \( 3x^2 + 2x - 4 = 0 \).
   Show the full solution

   \( a = 3 \), \( b = 2 \), \( c = -4 \). Discriminant: \( 4 - 4(3)(-4) = 4 + 48 = 52 \). \( \sqrt{52} = 2\sqrt{13} \). \( x = \dfrac{-2 \pm 2\sqrt{13}}{6} = \dfrac{-1 \pm \sqrt{13}}{3} \), after dividing numerator and denominator by 2. The reduction is only valid because every term shared the factor 2, including the \( -2 \). Canceling the 2 from the radical alone would be wrong. Check numerically: \( \sqrt{13} \approx 3.6056 \), so \( x \approx 0.8685 \). \( 3(0.7543) + 2(0.8685) - 4 \approx 2.263 + 1.737 - 4 = 0 \). Correct. \( x = \frac{-1 \pm \sqrt{13}}{3} \)

8. Solve \( x^2 = 6x - 4 \).
   Show the full solution

   Rearrange first. The formula needs zero on one side: \( x^2 - 6x + 4 = 0 \). So \( a = 1 \), \( b = -6 \), \( c = 4 \). Discriminant: \( 36 - 16 = 20 \). \( \sqrt{20} = 2\sqrt{5} \). \( x = \dfrac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5} \). Check exactly with \( 3 + \sqrt{5} \): \( (3 + \sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5} \). \( 6(3 + \sqrt{5}) - 4 = 18 + 6\sqrt{5} - 4 = 14 + 6\sqrt{5} \). Equal. Correct. A warning about skipping the rearrangement. Reading \( c \) as \( -4 \) from the unrearranged form would give a discriminant of 52 and completely wrong roots. \( x = 3 \pm \sqrt{5} \)

9. Explain why the quadratic formula is guaranteed to work on every quadratic equation.
   Show the full solution

   It is not an independent rule to be trusted. It is the recorded result of completing the square on the general equation, and every step of that derivation is valid for any coefficients. Step by step, what could have gone wrong and did not. Dividing by \( a \) is safe because \( a \ne 0 \) is part of what makes the equation quadratic. The completing constant \( \dfrac{b^2}{4a^2} \) always exists. Combining the right side over \( 4a^2 \) is ordinary fraction arithmetic. Taking the square root is the only step that can produce something unexpected, and it does so only when \( b^2 - 4ac \) is negative, in which case the roots are complex rather than nonexistent. What this buys. Since completing the square never fails, and the formula is that method executed once with letters, the formula never fails either. Factoring, by contrast, works only when convenient integers exist. The formula is the general method and factoring is the shortcut, not the other way round. When to use which. If the quadratic factors by inspection in a few seconds, factor. Otherwise use the formula. Completing the square is worth doing explicitly only when vertex form is what the question wants. It is completing the square performed once in general, and every step of that derivation holds for all coefficients

10. For what values of \( k \) does \( x^2 + kx + 9 = 0 \) have exactly one real solution? Find the solution in each case.
    Show the full solution

    Translate the condition. Exactly one real solution means the two roots coincide, which happens when the radical vanishes, that is when the discriminant is zero. Set up. \( a = 1 \), \( b = k \), \( c = 9 \). \( b^2 - 4ac = k^2 - 36 \). Set to zero: \( k^2 - 36 = 0 \), so \( k^2 = 36 \) and \( k = \pm 6 \). Two values, not one. Taking the square root of \( k^2 = 36 \) requires the \( \pm \), and losing the negative answer is the characteristic error here. Case \( k = 6 \). The equation is \( x^2 + 6x + 9 = 0 \), which is \( (x + 3)^2 = 0 \), so \( x = -3 \), a double root. Check: \( 9 - 18 + 9 = 0 \). Correct. Case \( k = -6 \). The equation is \( x^2 - 6x + 9 = 0 \), which is \( (x - 3)^2 = 0 \), so \( x = 3 \). Check: \( 9 - 18 + 9 = 0 \). Correct. The geometric reading. A discriminant of zero means the parabola touches the \( x \)-axis at exactly one point rather than crossing it, so the vertex sits on the axis. Confirming: for \( k = 6 \) the vertex is at \( x = -\dfrac{6}{2} = -3 \) with value \( 9 - 18 + 9 = 0 \). On the axis, as predicted. Why the two answers are mirror images. Changing the sign of \( b \) reflects the parabola in the \( y \)-axis, which moves the tangent point from \( -3 \) to 3 without changing the fact that it is tangent. The symmetry is structural, not a coincidence. \( k = 6 \) giving \( x = -3 \), or \( k = -6 \) giving \( x = 3 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Compute the discriminant of \( x^2 + 2x + 1 = 0 \).
   Show the full solution

   \( 4 - 4 \). 0

2. How many real roots does a quadratic with \( D = 0 \) have?
   Show the full solution

   One, repeated

3. Classify the roots of \( x^2 + x + 5 = 0 \).
   Show the full solution

   \( D = 1 - 20 = -19 \). Two complex roots

4. Classify the roots of \( 2x^2 - 7x + 3 = 0 \).
   Show the full solution

   \( D = 49 - 24 = 25 \), a perfect square. Two rational roots

5. What does \( D \lt 0 \) mean about the graph?
   Show the full solution

   It does not cross the \( x \)-axis

6. Classify the roots of \( 4x^2 - 12x + 9 = 0 \) and find them.
   Show the full solution

   \( D = 144 - 4(4)(9) = 144 - 144 = 0 \). One repeated real root, at \( x = -\dfrac{b}{2a} = \dfrac{12}{8} = \dfrac{3}{2} \). Confirming: \( 4x^2 - 12x + 9 = (2x - 3)^2 \), zero when \( 2x = 3 \). Check: \( 4\left( \dfrac{9}{4} \right) - 12\left( \dfrac{3}{2} \right) + 9 = 9 - 18 + 9 = 0 \). Correct. One repeated root, \( x = \frac{3}{2} \)

7. For what values of \( k \) does \( x^2 + kx + 4 = 0 \) have two distinct real roots?
   Show the full solution

   Two distinct real roots means \( D \gt 0 \): \( k^2 - 16 \gt 0 \). Factor: \( (k - 4)(k + 4) \gt 0 \), which is positive outside the roots. \( k \lt -4 \) or \( k \gt 4 \). Check \( k = 5 \): \( D = 25 - 16 = 9 \gt 0 \). Correct. Check \( k = 0 \): \( D = -16 \), so \( x^2 + 4 = 0 \) has no real roots. Correctly excluded. Check \( k = -5 \): \( D = 25 - 16 = 9 \gt 0 \). Correct, and this is why the negative branch belongs. \( k \lt -4 \) or \( k \gt 4 \)

8. A parabola \( y = x^2 + bx + 10 \) is tangent to the \( x \)-axis. Find \( b \) and the point of tangency.
   Show the full solution

   Tangent to the axis means exactly one root, so \( D = 0 \): \( b^2 - 40 = 0 \), giving \( b^2 = 40 \) and \( b = \pm 2\sqrt{10} \). Case \( b = 2\sqrt{10} \). The root is at \( x = -\dfrac{b}{2a} = -\sqrt{10} \). Point of tangency \( (-\sqrt{10},\, 0) \). Check: \( (-\sqrt{10})^2 + 2\sqrt{10}(-\sqrt{10}) + 10 = 10 - 20 + 10 = 0 \). Correct. Case \( b = -2\sqrt{10} \). The root is at \( x = \sqrt{10} \), and the same substitution gives \( 10 - 20 + 10 = 0 \). Correct. \( b = \pm 2\sqrt{10} \), touching at \( (\mp\sqrt{10},\, 0) \)

9. Explain why a quadratic with rational coefficients and irrational roots must have those roots in conjugate pairs.
   Show the full solution

   The two roots are \( \dfrac{-b + \sqrt{D}}{2a} \) and \( \dfrac{-b - \sqrt{D}}{2a} \). They differ only in the sign attached to the radical; everything else is identical. If the coefficients are rational then \( -b \), \( 2a \) and \( D \) are all rational, so the only irrational ingredient is \( \sqrt{D} \). Writing the roots as \( p \pm q\sqrt{n} \) with \( p \) and \( q \) rational shows the structure: the two roots share \( p \) and \( q \) and differ only in sign. A concrete case. \( x^2 - 6x + 7 = 0 \) has roots \( 3 + \sqrt{2} \) and \( 3 - \sqrt{2} \). Neither is rational, but their sum is 6 and their product is \( 9 - 2 = 7 \), both rational, matching \( -\dfrac{b}{a} \) and \( \dfrac{c}{a} \). Why the sum and product must be rational. They equal \( -\dfrac{b}{a} \) and \( \dfrac{c}{a} \), which are ratios of rational numbers. The irrational parts have to cancel in the sum and in the product, and the only way a pair of numbers of the form \( p \pm q\sqrt{n} \) can do both is by being exactly conjugate. Where this recurs. Lesson 2.6 proves the identical statement for complex roots, for the identical reason: the radical carries the whole difference between the two roots, so they are always conjugate. Unit 3 extends it to polynomials of any degree. The two roots differ only in the sign on \( \sqrt{D} \), so with rational coefficients they must be conjugates

10. For what values of \( m \) does the line \( y = mx + 1 \) intersect the parabola \( y = x^2 + 3 \) in exactly one point?
    Show the full solution

    Set them equal. An intersection is a point on both graphs, so \( mx + 1 = x^2 + 3 \). Rearrange to standard form. \( 0 = x^2 - mx + 2 \). Translate the condition. Exactly one intersection point means this quadratic has exactly one solution, so its discriminant is zero. The line is tangent to the parabola. \( D = (-m)^2 - 4(1)(2) = m^2 - 8 \). Set to zero: \( m^2 = 8 \), so \( m = \pm 2\sqrt{2} \). Find the point for \( m = 2\sqrt{2} \). The repeated root is \( x = \dfrac{m}{2} = \sqrt{2} \). Then \( y = x^2 + 3 = 2 + 3 = 5 \). Check with the line: \( y = 2\sqrt{2}\cdot\sqrt{2} + 1 = 4 + 1 = 5 \). Agrees, so \( (\sqrt{2},\, 5) \) is the tangent point. Find the point for \( m = -2\sqrt{2} \). By the mirror symmetry, the root is \( x = -\sqrt{2} \) and the point is \( (-\sqrt{2},\, 5) \). Check: parabola gives \( 2 + 3 = 5 \); line gives \( -2\sqrt{2}(-\sqrt{2}) + 1 = 5 \). Agrees. Sanity check on the whole setup. Try \( m = 0 \), a horizontal line \( y = 1 \). The parabola's minimum is 3, so they never meet, and indeed \( D = -8 \lt 0 \). Try \( m = 4 \), which is steeper than \( 2\sqrt{2} \approx 2.83 \): \( D = 16 - 8 = 8 \gt 0 \), two intersections. Both consistent. What the three cases mean geometrically. \( D \lt 0 \) is a line that misses the parabola, \( D = 0 \) is a tangent line, and \( D \gt 0 \) is a secant line cutting it twice. The discriminant of the combined equation classifies the line's relationship to the curve, which is the same role it played for the \( x \)-axis in the worked example. \( m = \pm 2\sqrt{2} \), touching at \( (\pm\sqrt{2},\, 5) \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Give answers in the form \( a + bi \).

1. Simplify \( i^3 \).
   Show the full solution

   \( i^2 \cdot i = -i \). \( -i \)

2. Simplify \( i^{12} \).
   Show the full solution

   12 is a multiple of 4. 1

3. Compute \( (5 + 2i) + (1 - 6i) \).
   Show the full solution

   \( 6 - 4i \)

4. Compute \( (2 + 3i)(2 - 3i) \).
   Show the full solution

   \( 4 + 9 \), by the conjugate rule. 13

5. Simplify \( \sqrt{-25} \).
   Show the full solution

   \( \sqrt{25}\sqrt{-1} \). \( 5i \)

6. Compute \( (4 - i)(3 + 2i) \).
   Show the full solution

   Expand: \( 12 + 8i - 3i - 2i^2 \). Combine: \( 12 + 5i - 2i^2 \). Substitute \( i^2 = -1 \): \( 12 + 5i + 2 = 14 + 5i \). \( 14 + 5i \)

7. Simplify \( i^{103} \).
   Show the full solution

   Divide 103 by 4: \( 103 = 4(25) + 3 \), remainder 3. \( i^{103} = i^3 = -i \). Reasoning: \( i^{103} = (i^4)^{25} \cdot i^3 = 1 \cdot (-i) = -i \). \( -i \)

8. Compute \( \dfrac{2 - i}{3 + i} \).
   Show the full solution

   Multiply by the conjugate \( 3 - i \) over itself. Numerator: \( (2 - i)(3 - i) = 6 - 2i - 3i + i^2 = 6 - 5i - 1 = 5 - 5i \). Denominator: \( (3 + i)(3 - i) = 9 + 1 = 10 \). \( \dfrac{5 - 5i}{10} = \dfrac{1}{2} - \dfrac{1}{2}i \). Check by multiplying back: \( \left( \dfrac{1}{2} - \dfrac{1}{2}i \right)(3 + i) = \dfrac{3}{2} + \dfrac{1}{2}i - \dfrac{3}{2}i - \dfrac{1}{2}i^2 = \dfrac{3}{2} - i + \dfrac{1}{2} = 2 - i \). Correct. \( \frac{1}{2} - \frac{1}{2}i \)

9. Explain why \( \sqrt{-4} \cdot \sqrt{-9} \) is not \( \sqrt{36} \).
   Show the full solution

   Compute it correctly first. \( \sqrt{-4} = 2i \) and \( \sqrt{-9} = 3i \). Their product is \( 2i \cdot 3i = 6i^2 = -6 \). The tempting wrong route. Applying \( \sqrt{a}\sqrt{b} = \sqrt{ab} \) would give \( \sqrt{(-4)(-9)} = \sqrt{36} = 6 \), the wrong sign entirely. Why the rule fails. The identity \( \sqrt{a}\sqrt{b} = \sqrt{ab} \) is proved for nonnegative \( a \) and \( b \), and its proof uses the fact that the principal square root is well defined on nonnegatives. With negative radicands that guarantee is gone, and the rule breaks. The safe procedure. Convert every negative radicand to \( i \) form before multiplying. Write \( \sqrt{-4} = 2i \) first, then multiply. Doing the conversion first and the multiplication second is always correct. A second case to confirm the danger. \( \sqrt{-1}\cdot\sqrt{-1} = i \cdot i = -1 \), whereas the misapplied rule gives \( \sqrt{1} = 1 \). The two answers differ in sign, and the correct one is \( -1 \) by the definition of \( i \). The radical product rule requires nonnegative radicands; convert to \( i \) form first, giving \( -6 \)

10. Show that \( 1 + 2i \) is a solution of \( x^2 - 2x + 5 = 0 \), find the other solution, and verify the sum and product of the roots.
    Show the full solution

    Substitute \( x = 1 + 2i \). \( (1 + 2i)^2 = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i \). \( -2(1 + 2i) = -2 - 4i \). Sum with the 5: \( (-3 + 4i) + (-2 - 4i) + 5 = (-3 - 2 + 5) + (4 - 4)i = 0 + 0i = 0 \). Verified. Find the other root. By the conjugate pair theorem of lesson 2.6, a real quadratic with one non-real root has its conjugate as the other: \( 1 - 2i \). Confirm by the quadratic formula: \( D = 4 - 20 = -16 \), so \( x = \dfrac{2 \pm \sqrt{-16}}{2} = \dfrac{2 \pm 4i}{2} = 1 \pm 2i \). Agrees. Verify the sum. It should be \( -\dfrac{b}{a} = 2 \). \( (1 + 2i) + (1 - 2i) = 2 \). The imaginary parts cancel, as conjugates must. Correct. Verify the product. It should be \( \dfrac{c}{a} = 5 \). \( (1 + 2i)(1 - 2i) = 1^2 + 2^2 = 1 + 4 = 5 \), by the conjugate rule. Correct. What the check demonstrates. Both the sum and the product came out real, even though neither root is. That is not luck: a quadratic with real coefficients has \( -\dfrac{b}{a} \) and \( \dfrac{c}{a} \) real by construction, so its roots must be arranged so that their imaginary parts cancel in the sum and annihilate in the product. Conjugate pairs are the only arrangement that does both, which is the content of the next lesson's theorem. Verified; the other root is \( 1 - 2i \), with sum 2 and product 5

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( \sqrt{-49} \).
   Show the full solution

   \( 7i \)

2. Solve \( x^2 + 16 = 0 \).
   Show the full solution

   \( x^2 = -16 \). \( x = \pm 4i \)

3. If \( 2 + 5i \) is a root of a real quadratic, what is the other root?
   Show the full solution

   \( 2 - 5i \)

4. Find the sum of \( 4 + 3i \) and its conjugate.
   Show the full solution

   8

5. Find the product of \( 4 + 3i \) and its conjugate.
   Show the full solution

   \( 16 + 9 \). 25

6. Solve \( x^2 - 6x + 10 = 0 \).
   Show the full solution

   \( D = 36 - 40 = -4 \), so \( \sqrt{-4} = 2i \). \( x = \dfrac{6 \pm 2i}{2} = 3 \pm i \). Check \( 3 + i \): \( (3 + i)^2 = 9 + 6i - 1 = 8 + 6i \). \( -6(3 + i) = -18 - 6i \). Total with 10: \( 8 + 6i - 18 - 6i + 10 = 0 \). Correct. \( x = 3 \pm i \)

7. Build a monic quadratic with real coefficients having \( -1 + 4i \) as a root.
   Show the full solution

   The conjugate \( -1 - 4i \) must also be a root. Sum: \( -2 \). Product: \( (-1)^2 + 4^2 = 1 + 16 = 17 \). \( x^2 - (-2)x + 17 = x^2 + 2x + 17 \). Check: \( D = 4 - 68 = -64 \), so \( x = \dfrac{-2 \pm 8i}{2} = -1 \pm 4i \). Correct. \( x^2 + 2x + 17 \)

8. Factor \( x^4 - 16 \) completely over the complex numbers.
   Show the full solution

   Difference of squares twice: \( x^4 - 16 = (x^2 - 4)(x^2 + 4) \). The first factors over the reals: \( (x - 2)(x + 2) \). The second has roots \( x^2 = -4 \), so \( x = \pm 2i \), giving \( (x - 2i)(x + 2i) \). Complete factorization: \( (x - 2)(x + 2)(x - 2i)(x + 2i) \). Check by expanding the complex pair: \( x^2 - 4i^2 = x^2 + 4 \). Correct, and \( (x^2 - 4)(x^2 + 4) = x^4 - 16 \). Correct. Four roots for a degree-four polynomial, two real and one conjugate pair, which is the fundamental theorem of algebra of lesson 3.6. \( (x-2)(x+2)(x-2i)(x+2i) \)

9. Explain why a real quadratic cannot have exactly one complex root.
   Show the full solution

   Suppose it had one non-real root and one real root. The conjugate pair theorem says the conjugate of the non-real root must also be a root, so that would be three roots in total for a quadratic. A quadratic has at most two roots, since a polynomial of degree \( n \) has at most \( n \) roots. The conjugate is not equal to the original, because a number equals its own conjugate only when its imaginary part is zero, that is only when it is real. So the three would be genuinely distinct. That is a contradiction, so the case is impossible. The complete list of possibilities. A real quadratic has either two real roots, counting a repeated root twice, or two non-real roots forming a conjugate pair. Nothing else. Confirmed by the discriminant. \( D \gt 0 \) gives two real, \( D = 0 \) gives one real repeated, \( D \lt 0 \) gives two complex. The negative case produces both roots at once, since the \( \pm \) applies to an imaginary quantity and yields two non-real values. There is no way to get one of each. The general version. For a real polynomial of any degree, non-real roots always come in pairs, so a real polynomial of odd degree must have at least one real root. That is why every cubic crosses the \( x \)-axis somewhere, which unit 3 uses. The conjugate would be a third root, which a quadratic cannot have

10. Show that \( (1 + i)^2 = 2i \), then use it to find all four solutions of \( x^4 = -4 \).
    Show the full solution

    Verify the given fact. \( (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i \). Confirmed. Set up the equation. Write \( x^4 = -4 \) as \( x^4 + 4 = 0 \). Factor as a difference of squares in disguise. Add and subtract \( 4x^2 \) to build a perfect square: \( x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2 = (x^2 + 2)^2 - (2x)^2 \). That is a difference of squares: \( = (x^2 + 2 - 2x)(x^2 + 2 + 2x) \). Solve each quadratic. First: \( x^2 - 2x + 2 = 0 \), with \( D = 4 - 8 = -4 \), so \( x = \dfrac{2 \pm 2i}{2} = 1 \pm i \). Second: \( x^2 + 2x + 2 = 0 \), with \( D = 4 - 8 = -4 \), so \( x = \dfrac{-2 \pm 2i}{2} = -1 \pm i \). The four solutions: \( 1 + i \), \( 1 - i \), \( -1 + i \), \( -1 - i \). Check one using the given fact. \( (1 + i)^4 = \big[(1 + i)^2\big]^2 = (2i)^2 = 4i^2 = -4 \). Correct. Check another. \( (-1 + i)^2 = 1 - 2i + i^2 = -2i \), so \( (-1 + i)^4 = (-2i)^2 = 4i^2 = -4 \). Correct. Note the structure. The four solutions form two conjugate pairs, \( 1 \pm i \) and \( -1 \pm i \), exactly as a real polynomial requires. They also sit symmetrically: each has real and imaginary parts of size 1, so all four lie at the same distance \( \sqrt{2} \) from the origin, spaced a quarter turn apart. That regular spacing is a general fact about the \( n \)th roots of any number, developed properly in Precalculus. Why the factoring trick worked. Adding and subtracting \( 4x^2 \) is completing the square on \( x^4 + 4 \) viewed as a quadratic in \( x^2 \), then using the difference of squares. It is the same technique as lesson 2.2 applied one level up, which is the substitution idea that lesson 3.5 develops. \( x = 1 + i,\; 1 - i,\; -1 + i,\; -1 - i \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( (x - 1)(x - 4) \lt 0 \).
   Show the full solution

   Negative between the roots. \( 1 \lt x \lt 4 \)

2. Solve \( (x + 2)(x - 5) \gt 0 \).
   Show the full solution

   Positive outside the roots. \( x \lt -2 \) or \( x \gt 5 \)

3. Solve \( x^2 - 9 \le 0 \).
   Show the full solution

   \( (x-3)(x+3) \le 0 \). \( -3 \le x \le 3 \)

4. Find the vertex of \( h(t) = -16t^2 + 32t \).
   Show the full solution

   \( t = -\dfrac{32}{-32} = 1 \), and \( h(1) = -16 + 32 = 16 \). \( (1, 16) \)

5. Solve \( x^2 + 4 \gt 0 \).
   Show the full solution

   No real roots, and the expression is always positive. All real numbers

6. Solve \( 2x^2 - 5x - 3 \ge 0 \).
   Show the full solution

   Factor: \( 2x^2 - 5x - 3 = (2x + 1)(x - 3) \). Check the factoring: \( 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3 \). Correct. Roots: \( x = -\dfrac{1}{2} \) and \( x = 3 \). The parabola opens up, so it is nonnegative outside the roots, inclusive: \( x \le -\dfrac{1}{2} \) or \( x \ge 3 \). Check \( x = 0 \): \( -3 \), which fails, and 0 is correctly between the roots. Check \( x = 4 \): \( 32 - 20 - 3 = 9 \ge 0 \). Correct. \( \left( -\infty, -\frac{1}{2} \right] \cup [3, \infty) \)

7. Solve \( x^2 + 2x + 5 \lt 0 \).
   Show the full solution

   Discriminant: \( 4 - 20 = -16 \), negative, so there are no real roots and the sign never changes. Test any value, say \( x = 0 \): the expression is 5, positive. Since it is positive somewhere and never changes sign, it is positive everywhere. The inequality asks for negative values. No solution Confirming with vertex form: \( x^2 + 2x + 1 + 4 = (x + 1)^2 + 4 \), whose minimum is 4. Never negative.

8. A rocket's height in meters after \( t \) seconds is \( h(t) = -5t^2 + 40t \). Find the maximum height, the landing time, and the interval when it is above 60 m.
   Show the full solution

   Maximum. \( t = -\dfrac{40}{2(-5)} = 4 \) seconds. \( h(4) = -80 + 160 = 80 \) m. Landing. \( -5t^2 + 40t = 0 \), so \( -5t(t - 8) = 0 \) and \( t = 0 \) or \( t = 8 \). \( t = 0 \) is the launch, so it lands at \( t = 8 \) seconds. Note the symmetry: the peak at \( t = 4 \) is exactly halfway. Correct. Above 60 m. \( -5t^2 + 40t \gt 60 \), so \( -5t^2 + 40t - 60 \gt 0 \). Divide by \( -5 \) and flip: \( t^2 - 8t + 12 \lt 0 \). Factor: \( (t - 2)(t - 6) \lt 0 \), so \( 2 \lt t \lt 6 \). Check \( t = 4 \): \( h = 80 \gt 60 \). Correct. Check \( t = 1 \): \( h = -5 + 40 = 35 \), not above 60. Correctly excluded. Max 80 m at 4 s; lands at 8 s; above 60 m from 2 s to 6 s

9. Explain why the sign of a quadratic can only change at its roots.
   Show the full solution

   A polynomial is a continuous function: its graph has no breaks or jumps. To go from a positive value to a negative one, a continuous graph must pass through zero somewhere in between. So a sign change forces a root. Contrapositively, on any interval containing no root, the sign cannot change, and testing a single point in that interval determines the sign on the whole of it. Why this makes the test-point method valid. Marking the roots cuts the number line into intervals, none of which contains a root in its interior. On each one the sign is constant, so one test value settles it. Without continuity the method would be unjustified. A contrast that shows the point. The function \( \dfrac{1}{x} \) changes sign at \( x = 0 \) without having a root there, because it is not continuous: it has a vertical asymptote. So for rational functions the sign can change at a root or at a value excluded from the domain, and lesson 4.6 handles that extra case. For polynomials, roots are the only possibility. A related consequence. The same continuity argument gives the intermediate value property used in unit 3: if a polynomial is negative at one point and positive at another, it has a root between them. That is how a root is located numerically when no exact method applies. Polynomials are continuous, so passing from positive to negative requires crossing zero

10. A company's profit from selling \( x \) units is \( P(x) = -0.5x^2 + 60x - 1000 \) dollars. Find the break-even points, the maximum profit, and the production range that keeps profit above $500.
    Show the full solution

    Break-even points. Profit is zero: \( -0.5x^2 + 60x - 1000 = 0 \). Multiply through by \( -2 \) to clear the decimal: \( x^2 - 120x + 2000 = 0 \). Discriminant: \( 14400 - 8000 = 6400 \), and \( \sqrt{6400} = 80 \). \( x = \dfrac{120 \pm 80}{2} \), giving \( x = 100 \) or \( x = 20 \). Check \( x = 20 \): \( -0.5(400) + 1200 - 1000 = -200 + 200 = 0 \). Correct. Check \( x = 100 \): \( -0.5(10000) + 6000 - 1000 = -5000 + 5000 = 0 \). Correct. Maximum profit. \( x = -\dfrac{60}{2(-0.5)} = 60 \) units. \( P(60) = -0.5(3600) + 3600 - 1000 = -1800 + 2600 = 800 \) dollars. Sanity check: 60 is the midpoint of 20 and 100, as the axis of symmetry must be. Profit above $500. \( -0.5x^2 + 60x - 1000 \gt 500 \), so \( -0.5x^2 + 60x - 1500 \gt 0 \). Multiply by \( -2 \) and flip: \( x^2 - 120x + 3000 \lt 0 \). Discriminant: \( 14400 - 12000 = 2400 \), and \( \sqrt{2400} = 20\sqrt{6} \approx 48.99 \). \( x = \dfrac{120 \pm 20\sqrt{6}}{2} = 60 \pm 10\sqrt{6} \), approximately 35.51 and 84.49. Negative between the roots: \( 35.51 \lt x \lt 84.49 \). Check \( x = 60 \): profit 800, above 500. Correct. Check \( x = 30 \): \( -450 + 1800 - 1000 = 350 \), below 500. Correctly excluded. Interpreting in context. Units sold are whole numbers, so the usable answer is 36 through 84 units. Verify the endpoints: \( P(36) = -648 + 2160 - 1000 = 512 \), above 500. And \( P(35) = -612.5 + 2100 - 1000 = 487.5 \), below. So 36 is genuinely the first qualifying value. At the top: \( P(84) = -3528 + 5040 - 1000 = 512 \), above 500, and \( P(85) = -3612.5 + 5100 - 1000 = 487.5 \), below. So 84 is the last. What the model assumes. It treats every unit produced as sold at a price that falls with volume, which is what the negative quadratic term encodes, and it assumes the $1000 fixed cost does not change with output. Beyond 100 units the model predicts losses, and beyond some point it would predict arbitrarily large losses, which no real firm would incur because it would stop producing. The model is trustworthy in the neighborhood of the data it was fitted to and not outside it. Break even at 20 and 100 units; maximum profit $800 at 60 units; above $500 for 36 to 84 units

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Give the degree of \( 4x^5 - x^2 + 9 \).
   Show the full solution

   5

2. Give the leading coefficient of \( -3x^4 + 2x \).
   Show the full solution

   \( -3 \)

3. Describe the end behavior of \( y = x^2 \).
   Show the full solution

   Both ends rise

4. Describe the end behavior of \( y = -x^3 \).
   Show the full solution

   Odd degree, negative leading coefficient. Rises left, falls right

5. Is \( \dfrac{1}{x} + 3 \) a polynomial?
   Show the full solution

   A variable in a denominator is not allowed. No

6. Give the degree, leading coefficient and end behavior of \( 7 - 2x^6 + x^3 \).
   Show the full solution

   Rewrite in standard form: \( -2x^6 + x^3 + 7 \). Degree 6, leading coefficient \( -2 \). Even degree with a negative leading coefficient, so both ends fall. Check: \( f(10) = -2{,}000{,}000 + 1000 + 7 \), large and negative. And \( f(-10) = -2{,}000{,}000 - 1000 + 7 \), also large and negative. Correct. Degree 6, leading coefficient \( -2 \), both ends fall

7. A polynomial of degree 5 has a negative leading coefficient. How many real roots could it have?
   Show the full solution

   At most 5, since the degree bounds the number of roots. At least 1, because odd degree forces the graph to run from \( +\infty \) to \( -\infty \) and therefore to cross the axis. Non-real roots come in conjugate pairs, so removing them from 5 removes an even number each time. The possible counts of real roots, with multiplicity, are 1, 3 or 5. The leading coefficient being negative affects which end goes which way but not the counting. 1, 3 or 5

8. Two polynomials have degree 4 and degree 3. What are the possible degrees of their sum and of their product?
   Show the full solution

   Product. Multiplying the leading terms gives a term of degree \( 4 + 3 = 7 \), and nothing can cancel it since no other product of terms reaches degree 7. The product has degree exactly 7. Sum. The degree-4 term has nothing to cancel against, because the second polynomial has no degree-4 term at all. So the sum has degree exactly 4. When cancellation is possible. Had both polynomials been degree 4, the sum could drop: \( (x^4 + x) + (-x^4 + 1) = x + 1 \), degree 1. Equal degrees are the only case where the sum's degree is uncertain, and then it can be anything from 0 up to that degree, or the zero polynomial. Product degree 7; sum degree 4

9. Explain why end behavior depends only on the leading term.
   Show the full solution

   Factor the leading power out of the whole polynomial. For \( a_n x^n + a_{n-1}x^{n-1} + \cdots + a_0 \) this gives \[ x^n\left( a_n + \frac{a_{n-1}}{x} + \frac{a_{n-2}}{x^2} + \cdots + \frac{a_0}{x^n} \right) \] Every term inside the bracket except \( a_n \) has a positive power of \( x \) in its denominator, so each approaches zero as \( \left| x \right| \) grows. The bracket approaches \( a_n \), and the whole expression behaves like \( a_n x^n \). What decides the picture. The parity of \( n \) decides whether the two ends agree: an even power turns a large negative input into a large positive output, so both ends match, while an odd power preserves sign and the ends oppose. The sign of \( a_n \) then flips both ends or neither. The limits of the statement. End behavior says nothing about the middle. Two polynomials with identical end behavior can look completely different near the origin, with different numbers of turns and roots. End behavior is the first thing to establish when sketching and never the last. Dividing by the leading power sends every other term to zero, leaving the leading coefficient

10. Sketch-describe a possible graph of a degree-4 polynomial with positive leading coefficient, exactly two real roots, and a negative \( y \)-intercept. Give a formula.
    Show the full solution

    Work out what the description forces. Degree 4 with a positive leading coefficient means both ends rise. Exactly two real roots means the graph crosses or touches the axis at exactly two places. A negative \( y \)-intercept means it is below the axis at \( x = 0 \). Reconcile them. If the graph rises at both ends and dips below the axis in the middle, it must cross the axis twice, once coming down and once going back up. Between those crossings it stays below, which accommodates the negative \( y \)-intercept provided 0 lies between the two roots. Account for the other two roots. Degree 4 requires four roots counted with multiplicity, and only two are real. So the other two form a complex conjugate pair, contributing an irreducible quadratic factor. Build a formula. Take real roots at \( -1 \) and 2, and an irreducible quadratic factor such as \( x^2 + 1 \), which has no real roots since its discriminant is \( -4 \): \( f(x) = (x + 1)(x - 2)(x^2 + 1) \). Check the leading coefficient. Multiplying the leading terms gives \( x \cdot x \cdot x^2 = x^4 \), so the coefficient is \( +1 \). Positive, as required. Check the \( y \)-intercept. \( f(0) = (1)(-2)(1) = -2 \). Negative, as required. Check the roots. \( f(-1) = 0 \) and \( f(2) = 0 \), and \( x^2 + 1 \) contributes no real roots. Exactly two. Correct. Check the shape between them. At \( x = 1 \): \( (2)(-1)(2) = -4 \), below the axis, consistent with the dip. At \( x = 3 \): \( (4)(1)(10) = 40 \), above. At \( x = -2 \): \( (-1)(-4)(5) = 20 \), above. So the graph is above, dips below between \( -1 \) and 2, and returns above. Matches the description. Note that the answer is not unique. Any positive multiple works, as does any other pair of real roots straddling zero and any irreducible quadratic. The question asked for a possible graph, and demonstrating that the constraints are consistent is the substance of the answer. \( f(x) = (x+1)(x-2)(x^2+1) \), with both ends rising, crossings at \( -1 \) and 2, and \( f(0) = -2 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( (2x + 3) + (5x - 1) \).
   Show the full solution

   \( 7x + 2 \)

2. Simplify \( (4x^2 - x) - (x^2 + 3x) \).
   Show the full solution

   \( 4x^2 - x - x^2 - 3x \). \( 3x^2 - 4x \)

3. Expand \( (x + 5)(x - 5) \).
   Show the full solution

   Difference of squares. \( x^2 - 25 \)

4. Expand \( (2x - 3)^2 \).
   Show the full solution

   \( 4x^2 - 12x + 9 \). \( 4x^2 - 12x + 9 \)

5. What is the degree of the product of a degree-3 and a degree-5 polynomial?
   Show the full solution

   8

6. Expand \( (x + 2)(x^2 - 3x + 4) \).
   Show the full solution

   Six products: \( x \) against each: \( x^3 \), \( -3x^2 \), \( 4x \). \( 2 \) against each: \( 2x^2 \), \( -6x \), \( 8 \). Collect: \( x^3 + (-3 + 2)x^2 + (4 - 6)x + 8 = x^3 - x^2 - 2x + 8 \). Check at \( x = 1 \): \( (3)(2) = 6 \), and \( 1 - 1 - 2 + 8 = 6 \). Agrees. \( x^3 - x^2 - 2x + 8 \)

7. Expand \( (x - 1)^3 \).
   Show the full solution

   First square: \( (x - 1)^2 = x^2 - 2x + 1 \). Then multiply by \( (x - 1) \): \( x \) against each: \( x^3 \), \( -2x^2 \), \( x \). \( -1 \) against each: \( -x^2 \), \( 2x \), \( -1 \). Collect: \( x^3 - 3x^2 + 3x - 1 \). Check at \( x = 2 \): \( (1)^3 = 1 \), and \( 8 - 12 + 6 - 1 = 1 \). Agrees. The coefficients 1, 3, 3, 1 are the fourth row of Pascal's triangle, with alternating signs from the minus. Lesson 8.7 makes that general. \( x^3 - 3x^2 + 3x - 1 \)

8. Expand \( (x^2 + 2x - 1)(x^2 - 2x + 1) \) and simplify.
   Show the full solution

   Spot the structure first. The second factor is \( (x - 1)^2 \), and the first can be written as \( x^2 + (2x - 1) \), which does not obviously help. Expand directly instead, but notice that grouping will pay off: write the first as \( x^2 + (2x - 1) \) and the second as \( x^2 - (2x - 1) \). That is a difference of squares with \( a = x^2 \) and \( b = 2x - 1 \): \( a^2 - b^2 = x^4 - (2x - 1)^2 = x^4 - (4x^2 - 4x + 1) = x^4 - 4x^2 + 4x - 1 \). Check by direct expansion at \( x = 2 \). First factor: \( 4 + 4 - 1 = 7 \). Second: \( 4 - 4 + 1 = 1 \). Product: 7. Answer: \( 16 - 16 + 8 - 1 = 7 \). Agrees. Check at \( x = 3 \). First: \( 9 + 6 - 1 = 14 \). Second: \( 9 - 6 + 1 = 4 \). Product: 56. Answer: \( 81 - 36 + 12 - 1 = 56 \). Agrees. \( x^4 - 4x^2 + 4x - 1 \)

9. Explain why checking an expansion at \( x = 1 \) alone is not sufficient.
   Show the full solution

   Substituting \( x = 1 \) makes every power of \( x \) equal 1, so the check reduces to comparing the sum of the coefficients on each side. Many different wrong answers have the same coefficient sum as the right one. A concrete failure. Suppose the correct expansion is \( x^2 + 3x + 2 \), with coefficient sum 6. The wrong answer \( x^2 + 2x + 3 \) also sums to 6, so the \( x = 1 \) check passes on an answer with two coefficients swapped. At \( x = 2 \), the correct one gives \( 4 + 6 + 2 = 12 \) and the wrong one gives \( 4 + 4 + 3 = 11 \). The second check separates them. Why \( x = 0 \) is also weak. It tests only the constant term, since every other term vanishes. It is a useful quick check of the constant and nothing more. How many checks are enough. Two polynomials of degree \( n \) that agree at \( n + 1 \) distinct values are identical, so a degree-4 expansion is fully verified by five well-chosen values. In practice two or three catch essentially every arithmetic slip, and \( x = 2 \) is a good default because it weights the higher powers heavily. At \( x = 1 \) the check compares only the sums of the coefficients, which many wrong answers also match

10. A box is made from a 20 cm by 30 cm sheet by cutting squares of side \( x \) from each corner and folding up. Write its volume as a polynomial, give the valid domain, and find the volume at \( x = 4 \).
    Show the full solution

    Work out the dimensions. Cutting a square of side \( x \) from each corner removes \( x \) from both ends of each dimension. Base length: \( 30 - 2x \). Base width: \( 20 - 2x \). Height: \( x \), the folded-up flap. Write the volume. \( V(x) = x(30 - 2x)(20 - 2x) \). Expand. Multiply the two brackets first: \( (30 - 2x)(20 - 2x) = 600 - 60x - 40x + 4x^2 = 4x^2 - 100x + 600 \). Then multiply by \( x \): \( V(x) = 4x^3 - 100x^2 + 600x \). Check the degree. Three linear factors give degree 3. Correct, and a volume built from three lengths should be cubic. Find the domain the situation allows. The height must be positive, so \( x \gt 0 \). Both base dimensions must be positive, so \( 30 - 2x \gt 0 \) giving \( x \lt 15 \), and \( 20 - 2x \gt 0 \) giving \( x \lt 10 \). The binding constraint is the smaller one: \( 0 \lt x \lt 10 \). Why the narrower constraint wins. The shorter side runs out first. A cut of 12 cm would satisfy \( x \lt 15 \) but would consume more than the entire 20 cm width, which is physically impossible. Evaluate at \( x = 4 \). From the factored form: \( 4(30 - 8)(20 - 8) = 4(22)(12) = 1056 \) cubic centimeters. From the expanded form: \( 4(64) - 100(16) + 600(4) = 256 - 1600 + 2400 = 1056 \). Both agree, which checks the expansion. A sanity check on the size. The original sheet is 600 square centimeters. A box roughly 22 by 12 by 4 holding about a liter is plausible for that amount of material. \( V(x) = 4x^3 - 100x^2 + 600x \) on \( 0 \lt x \lt 10 \); \( V(4) = 1056 \) cubic cm

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What value of \( c \) does synthetic division use for the divisor \( x - 5 \)?
   Show the full solution

   5

2. What value for the divisor \( x + 3 \)?
   Show the full solution

   \( x + 3 = x - (-3) \). \( -3 \)

3. List the coefficients to use for \( x^3 + 5 \).
   Show the full solution

   Insert zeros for the missing degrees. \( 1, 0, 0, 5 \)

4. Divide \( x^2 + 5x + 6 \) by \( x + 2 \).
   Show the full solution

   \( c = -2 \): bring down 1; \( 5 - 2 = 3 \); \( 6 - 6 = 0 \). \( x + 3 \), remainder 0

5. If the remainder is zero, what does that mean?
   Show the full solution

   The divisor is a factor

6. Divide \( 3x^3 - 2x^2 + 4x - 1 \) by \( x - 1 \).
   Show the full solution

   \( c = 1 \), coefficients \( 3, -2, 4, -1 \). Bring down 3. \( 3 \times 1 = 3 \); \( -2 + 3 = 1 \). \( 1 \times 1 = 1 \); \( 4 + 1 = 5 \). \( 5 \times 1 = 5 \); \( -1 + 5 = 4 \). Quotient \( 3x^2 + x + 5 \), remainder 4. Check: \( (x-1)(3x^2 + x + 5) + 4 = 3x^3 + x^2 + 5x - 3x^2 - x - 5 + 4 = 3x^3 - 2x^2 + 4x - 1 \). Correct. \( 3x^2 + x + 5 \), remainder 4

7. Divide \( x^4 - 16 \) by \( x - 2 \).
   Show the full solution

   Coefficients with zeros: \( 1, 0, 0, 0, -16 \). \( c = 2 \). Bring down 1. \( 2 \); \( 0 + 2 = 2 \). \( 4 \); \( 0 + 4 = 4 \). \( 8 \); \( 0 + 8 = 8 \). \( 16 \); \( -16 + 16 = 0 \). Quotient \( x^3 + 2x^2 + 4x + 8 \), remainder 0. Check: \( (x - 2)(x^3 + 2x^2 + 4x + 8) = x^4 + 2x^3 + 4x^2 + 8x - 2x^3 - 4x^2 - 8x - 16 = x^4 - 16 \). Correct. Three zeros had to be inserted here; omitting them is the classic failure. \( x^3 + 2x^2 + 4x + 8 \), remainder 0

8. Divide \( 2x^3 + x^2 - 8x + 5 \) by \( 2x - 1 \).
   Show the full solution

   The divisor is not monic, so factor it: \( 2x - 1 = 2\left( x - \dfrac{1}{2} \right) \). Divide synthetically by \( x - \dfrac{1}{2} \), with \( c = \dfrac{1}{2} \) and coefficients \( 2, 1, -8, 5 \). Bring down 2. \( 2 \times \dfrac{1}{2} = 1 \); \( 1 + 1 = 2 \). \( 2 \times \dfrac{1}{2} = 1 \); \( -8 + 1 = -7 \). \( -7 \times \dfrac{1}{2} = -\dfrac{7}{2} \); \( 5 - \dfrac{7}{2} = \dfrac{3}{2} \). So \( P = \left( x - \dfrac{1}{2} \right)(2x^2 + 2x - 7) + \dfrac{3}{2} \). Now account for the factor of 2. Since \( 2x - 1 = 2\left( x - \dfrac{1}{2} \right) \), divide the quotient by 2: \( P = (2x - 1)\left( x^2 + x - \dfrac{7}{2} \right) + \dfrac{3}{2} \). Quotient \( x^2 + x - \dfrac{7}{2} \), remainder \( \dfrac{3}{2} \). Check at \( x = 1 \): \( P(1) = 2 + 1 - 8 + 5 = 0 \). And \( (1)\left( 1 + 1 - \dfrac{7}{2} \right) + \dfrac{3}{2} = -\dfrac{3}{2} + \dfrac{3}{2} = 0 \). Correct. The remainder was not divided by 2, because only the quotient absorbs the factor. \( x^2 + x - \frac{7}{2} \), remainder \( \frac{3}{2} \)

9. Explain why the quotient has degree exactly one less than the dividend when dividing by a linear factor.
   Show the full solution

   Write the division as \( P = DQ + R \), where \( D \) is the linear divisor. Degrees add under multiplication, so \( DQ \) has degree \( 1 + \deg Q \). The remainder \( R \) has degree less than 1, so it is a constant and cannot affect the leading term. Therefore \( \deg P = 1 + \deg Q \), giving \( \deg Q = \deg P - 1 \). The bookkeeping consequence. Dividing a cubic by a linear factor must give a quadratic quotient. If synthetic division produces three numbers before the remainder, that is a quadratic and the count is right; two numbers would mean a coefficient was skipped, almost always a missing zero. Why the remainder is a constant. The algorithm stops when the remainder's degree drops below the divisor's. With a linear divisor that means degree 0 or the zero polynomial, so the remainder is always a single number. That is what makes the remainder theorem of lesson 3.4 possible: a constant remainder can be equal to a function value. The general version. Dividing by a divisor of degree \( d \) gives a quotient of degree \( \deg P - d \) and a remainder of degree at most \( d - 1 \). Dividing a quartic by a quadratic gives a quadratic quotient and a remainder that may be linear. Degrees add in \( DQ \), and the constant remainder cannot change the leading term, so \( \deg Q = \deg P - 1 \)

10. When \( P(x) = x^3 + kx^2 - 4x + 6 \) is divided by \( x - 3 \), the remainder is 12. Find \( k \), then factor \( P \) completely.
    Show the full solution

    Use synthetic division with \( c = 3 \). Coefficients \( 1, k, -4, 6 \). Bring down 1. \( 1 \times 3 = 3 \); \( k + 3 \). \( (k + 3) \times 3 = 3k + 9 \); \( -4 + 3k + 9 = 3k + 5 \). \( (3k + 5) \times 3 = 9k + 15 \); \( 6 + 9k + 15 = 9k + 21 \). Set the remainder equal to 12. \( 9k + 21 = 12 \), so \( 9k = -9 \) and \( k = -1 \). Check with the remainder theorem instead. \( P(3) = 27 + 9k - 12 + 6 = 9k + 21 \). Same expression, confirming the synthetic work. Setting it to 12 gives \( k = -1 \). Agrees. Write out \( P \). \( P(x) = x^3 - x^2 - 4x + 6 \). Find a root. Try the divisors of 6. \( P(1) = 1 - 1 - 4 + 6 = 2 \), not a root. \( P(-1) = -1 - 1 + 4 + 6 = 8 \), no. \( P(2) = 8 - 4 - 8 + 6 = 2 \), no. \( P(-2) = -8 - 4 + 8 + 6 = 2 \), no. \( P(3) = 27 - 9 - 12 + 6 = 12 \), no, as expected. \( P(-3) = -27 - 9 + 12 + 6 = -18 \), no. No rational roots exist. The rational root theorem of lesson 3.6 says any rational root must be a divisor of 6, and all eight candidates fail. So \( P \) does not factor over the rationals. It still has a real root, since it has odd degree, but that root is irrational. Numerically \( P(2.5) = 15.625 - 6.25 - 10 + 6 = 5.375 \) and \( P(1.5) = 3.375 - 2.25 - 6 + 6 = 1.125 \), while \( P(1) = 2 \) and \( P(1.2) = 1.728 - 1.44 - 4.8 + 6 = 1.488 \). All positive. Trying negative values: \( P(-2.5) = -15.625 - 6.25 + 10 + 6 = -5.875 \), and \( P(-2) = 2 \). So a root lies between \( -2.5 \) and \( -2 \). The honest answer. \( P(x) = x^3 - x^2 - 4x + 6 \) has one real irrational root near \( -2.2 \) and two complex roots, and it cannot be factored further over the rationals. Reporting that, rather than forcing a factorization, is the correct conclusion. \( k = -1 \); \( P \) has no rational roots and does not factor over the rationals

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the remainder when \( x^2 + 3x - 4 \) is divided by \( x - 1 \).
   Show the full solution

   \( P(1) = 1 + 3 - 4 \). 0

2. Is \( x - 1 \) a factor of that polynomial?
   Show the full solution

   The remainder is zero. Yes

3. Find the remainder when \( x^3 + 2x \) is divided by \( x - 2 \).
   Show the full solution

   \( 8 + 4 \). 12

4. What value do you substitute to test the factor \( x + 4 \)?
   Show the full solution

   \( x + 4 = 0 \) gives \( x = -4 \). \( -4 \)

5. If \( P(5) = 0 \), name a factor of \( P \).
   Show the full solution

   \( x - 5 \)

6. Find the remainder when \( 3x^4 - 2x^2 + 1 \) is divided by \( x + 1 \).
   Show the full solution

   \( c = -1 \). \( P(-1) = 3(1) - 2(1) + 1 = 3 - 2 + 1 = 2 \). Note that \( (-1)^4 = 1 \) and \( (-1)^2 = 1 \), both positive. 2

7. Show that \( x + 3 \) is a factor of \( x^3 + 2x^2 - 5x - 6 \) and factor completely.
   Show the full solution

   Test \( c = -3 \): \( -27 + 18 + 15 - 6 = 0 \). It is a factor. Synthetic division with \( -3 \) on \( 1, 2, -5, -6 \): bring down 1; \( -3 \), \( 2 - 3 = -1 \); \( 3 \), \( -5 + 3 = -2 \); \( 6 \), \( -6 + 6 = 0 \). Quotient \( x^2 - x - 2 = (x - 2)(x + 1) \). Complete factorization: \( (x + 3)(x - 2)(x + 1) \). Check the roots: \( P(2) = 8 + 8 - 10 - 6 = 0 \). Correct. \( P(-1) = -1 + 2 + 5 - 6 = 0 \). Correct. \( (x+3)(x-2)(x+1) \)

8. Find \( k \) so that \( x - 2 \) is a factor of \( x^3 + kx^2 - 7x + 2 \).
   Show the full solution

   By the factor theorem, \( x - 2 \) is a factor exactly when \( P(2) = 0 \). \( P(2) = 8 + 4k - 14 + 2 = 4k - 4 \). Set to zero: \( 4k = 4 \), so \( k = 1 \). Check: with \( k = 1 \), \( P(x) = x^3 + x^2 - 7x + 2 \) and \( P(2) = 8 + 4 - 14 + 2 = 0 \). Correct. Factoring further: synthetic division with 2 on \( 1, 1, -7, 2 \) gives \( 1, 3, -1, 0 \), so \( P(x) = (x - 2)(x^2 + 3x - 1) \). The quadratic has discriminant \( 9 + 4 = 13 \), so its roots are \( \dfrac{-3 \pm \sqrt{13}}{2} \), irrational. \( k = 1 \)

9. Explain why the remainder theorem makes testing candidate roots faster than dividing.
   Show the full solution

   Dividing produces both a quotient and a remainder, and the algorithm has to be run in full to get either. Evaluating produces only the remainder, but when the question is "is this a root", the remainder is the only thing wanted. The cost comparison. Testing \( P(c) \) on a cubic is three multiplications and three additions, done mentally for small \( c \). Synthetic division is the same arithmetic but with the intermediate results recorded, and long division is considerably more writing. The practical strategy this supports. Lesson 3.6 lists many candidate roots, often a dozen or more. Evaluating each is quick, and most fail. Only once a candidate succeeds is division performed, and then it is performed once rather than a dozen times. A refinement worth knowing. Evaluation is fastest in nested form. Writing \( 2x^3 + 3x^2 - 5x + 7 \) as \( ((2x + 3)x - 5)x + 7 \) needs three multiplications instead of six, and it is exactly the arithmetic synthetic division performs. So the two methods are the same computation, with division additionally recording the partial results that form the quotient. Evaluation gives the remainder alone, which is all a root test needs, and the quotient is only computed once a root is confirmed

10. A cubic \( P \) satisfies \( P(1) = 0 \), \( P(-2) = 0 \), \( P(3) = 0 \) and \( P(0) = 12 \). Find \( P \).
    Show the full solution

    Convert the roots to factors. By the factor theorem, each root gives a linear factor: \( P(1) = 0 \) gives the factor \( x - 1 \). \( P(-2) = 0 \) gives \( x + 2 \). \( P(3) = 0 \) gives \( x - 3 \). Account for the degree. Three linear factors multiply to a cubic, and \( P \) is a cubic, so there is no room for further factors. The only freedom left is a constant multiplier: \( P(x) = a(x - 1)(x + 2)(x - 3) \). Use the fourth condition to find \( a \). \( P(0) = a(-1)(2)(-3) = 6a \). Setting \( 6a = 12 \) gives \( a = 2 \). Write the answer. \( P(x) = 2(x - 1)(x + 2)(x - 3) \). Expand to standard form. First \( (x - 1)(x + 2) = x^2 + x - 2 \). Then \( (x^2 + x - 2)(x - 3) = x^3 - 3x^2 + x^2 - 3x - 2x + 6 = x^3 - 2x^2 - 5x + 6 \). Multiply by 2: \( P(x) = 2x^3 - 4x^2 - 10x + 12 \). Check all four conditions. \( P(1) = 2 - 4 - 10 + 12 = 0 \). Correct. \( P(-2) = -16 - 16 + 20 + 12 = 0 \). Correct. \( P(3) = 54 - 36 - 30 + 12 = 0 \). Correct. \( P(0) = 12 \). Correct. Why the answer is unique. Three roots fixed the factors up to a constant, and the fourth condition fixed the constant. In general a polynomial of degree \( n \) is determined by \( n + 1 \) conditions, and here four conditions determined a cubic exactly. Had only the three roots been given, every multiple \( a(x-1)(x+2)(x-3) \) would have qualified, so the value at a non-root was genuinely necessary. \( P(x) = 2(x-1)(x+2)(x-3) = 2x^3 - 4x^2 - 10x + 12 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Factor completely over the reals.

1. Factor \( x^2 - 49 \).
   Show the full solution

   \( (x-7)(x+7) \)

2. Factor \( x^3 + 8 \).
   Show the full solution

   Sum of cubes with \( a = x \), \( b = 2 \). \( (x+2)(x^2 - 2x + 4) \)

3. Factor \( 2x^3 + 6x^2 \).
   Show the full solution

   Common factor first. \( 2x^2(x + 3) \)

4. Factor \( x^3 - 27 \).
   Show the full solution

   \( (x-3)(x^2 + 3x + 9) \)

5. Does \( x^2 + 16 \) factor over the reals?
   Show the full solution

   A sum of squares does not. No

6. Factor \( x^3 - 2x^2 - 9x + 18 \) by grouping.
   Show the full solution

   Pair them: \( (x^3 - 2x^2) + (-9x + 18) \). First pair: \( x^2(x - 2) \). Second pair: \( -9(x - 2) \). Common bracket: \( (x - 2)(x^2 - 9) \). The second factor is a difference of squares: \( (x - 2)(x - 3)(x + 3) \). Check at \( x = 1 \): the original gives \( 1 - 2 - 9 + 18 = 8 \), and \( (-1)(-2)(4) = 8 \). Agrees. \( (x-2)(x-3)(x+3) \)

7. Factor \( x^4 - 5x^2 + 4 \).
   Show the full solution

   Quadratic form with \( u = x^2 \): \( u^2 - 5u + 4 = (u - 1)(u - 4) \). Substituting back: \( (x^2 - 1)(x^2 - 4) \). Both are differences of squares: \( (x - 1)(x + 1)(x - 2)(x + 2) \). Check at \( x = 3 \): the original gives \( 81 - 45 + 4 = 40 \), and \( (2)(4)(1)(5) = 40 \). Agrees. \( (x-1)(x+1)(x-2)(x+2) \)

8. Factor \( 16x^4 - 81 \).
   Show the full solution

   Difference of squares with \( a = 4x^2 \) and \( b = 9 \): \( (4x^2 - 9)(4x^2 + 9) \). The first is again a difference of squares, with \( a = 2x \) and \( b = 3 \): \( (2x - 3)(2x + 3) \). The second is a sum of squares and does not factor over the reals. \( (2x - 3)(2x + 3)(4x^2 + 9) \). Check at \( x = 1 \): the original gives \( 16 - 81 = -65 \), and \( (-1)(5)(13) = -65 \). Agrees. \( (2x-3)(2x+3)(4x^2+9) \)

9. Explain why a sum of squares does not factor over the reals but a sum of cubes does.
   Show the full solution

   A real factorization into linear factors would exhibit real roots, so the question is whether real roots exist. Sum of squares. \( a^2 + b^2 = 0 \) with real numbers requires both \( a = 0 \) and \( b = 0 \), since a sum of two nonnegative quantities is zero only when both are. So \( x^2 + k \) with \( k \gt 0 \) has no real root, and no real linear factor. It is irreducible over the reals. Sum of cubes. \( x^3 + k \) does have a real root, namely \( x = -\sqrt[3]{k} \), because odd roots of negatives exist. So a real linear factor exists, and it is \( x + \sqrt[3]{k} \). Dividing it out leaves the quadratic \( x^2 - \sqrt[3]{k}\,x + \sqrt[3]{k^2} \), which is the second factor in the formula. The underlying reason. It is the same parity distinction that runs through the course: even powers destroy sign information and odd ones preserve it. Squaring makes everything nonnegative, so a sum of squares can never reach zero without both parts vanishing. Cubing does not, so a cube can be negative and a sum of cubes can vanish. A check on the quadratic factor. In \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), the quadratic \( a^2 - ab + b^2 \), viewed in \( a \), has discriminant \( b^2 - 4b^2 = -3b^2 \), negative unless \( b = 0 \). So it is irreducible over the reals, confirming that the given formula is already complete. A sum of squares has no real root because squares are nonnegative; a sum of cubes does, because odd powers preserve sign

10. Factor \( x^6 - 64 \) completely over the reals, two different ways, and reconcile the results.
    Show the full solution

    First route: treat it as a difference of squares. \( x^6 - 64 = (x^3)^2 - 8^2 = (x^3 - 8)(x^3 + 8) \). Now each factor is a difference or sum of cubes: \( x^3 - 8 = (x - 2)(x^2 + 2x + 4) \). \( x^3 + 8 = (x + 2)(x^2 - 2x + 4) \). Result: \( (x - 2)(x + 2)(x^2 + 2x + 4)(x^2 - 2x + 4) \). Second route: treat it as a difference of cubes. \( x^6 - 64 = (x^2)^3 - 4^3 = (x^2 - 4)(x^4 + 4x^2 + 16) \). The first factor is a difference of squares: \( (x - 2)(x + 2) \). The quartic looks irreducible but is not. Add and subtract \( 4x^2 \): \( x^4 + 4x^2 + 16 = x^4 + 8x^2 + 16 - 4x^2 = (x^2 + 4)^2 - (2x)^2 \). That is a difference of squares: \( = (x^2 + 4 - 2x)(x^2 + 4 + 2x) = (x^2 - 2x + 4)(x^2 + 2x + 4) \). Result: \( (x - 2)(x + 2)(x^2 - 2x + 4)(x^2 + 2x + 4) \). Reconcile. The two results are identical, with the quadratic factors in the opposite order. Multiplication is commutative, so they are the same factorization. Check the quadratics are irreducible. \( x^2 + 2x + 4 \) has discriminant \( 4 - 16 = -12 \), negative. \( x^2 - 2x + 4 \) has discriminant \( 4 - 16 = -12 \), negative. Neither factors over the reals, so the factorization is complete. Verify numerically at \( x = 3 \). Original: \( 729 - 64 = 665 \). Factored: \( (1)(5)(9 - 6 + 4)(9 + 6 + 4) = (1)(5)(7)(19) = 665 \). Agrees. Verify at \( x = 1 \). Original: \( 1 - 64 = -63 \). Factored: \( (-1)(3)(1 - 2 + 4)(1 + 2 + 4) = (-1)(3)(3)(7) = -63 \). Agrees. Why the second route needed a trick. Choosing the cubes route first produced a quartic that no listed technique handles directly. The add-and-subtract move that rescued it is the same completing-the-square idea from lesson 2.2, used to manufacture a difference of squares. It is worth knowing precisely because it rescues expressions of the form \( x^4 + ax^2 + b^2 \) that appear irreducible. The practical lesson. When more than one technique applies, the order of application can make the work much easier or much harder, but the final complete factorization is unique up to the order of the factors and constant multiples. That uniqueness is the polynomial version of prime factorization for integers. \( (x-2)(x+2)(x^2-2x+4)(x^2+2x+4) \) by either route

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. List the candidate rational roots of \( x^3 + 2x - 5 \).
   Show the full solution

   Constant 5, leading coefficient 1. \( \pm 1, \pm 5 \)

2. How many complex roots does a degree-6 polynomial have?
   Show the full solution

   Counted with multiplicity. 6

3. Give the roots of \( (x - 4)^3 \) with multiplicity.
   Show the full solution

   \( x = 4 \), multiplicity 3

4. Is 2 a root of \( x^3 - 3x - 2 \)?
   Show the full solution

   \( 8 - 6 - 2 = 0 \). Yes

5. Does the rational root theorem guarantee a rational root exists?
   Show the full solution

   No, it only restricts where one could be

6. List the candidate rational roots of \( 3x^3 - x + 4 \).
   Show the full solution

   Divisors of the constant 4: \( 1, 2, 4 \). Divisors of the leading coefficient 3: \( 1, 3 \). Candidates: \( \pm 1, \pm 2, \pm 4, \pm\dfrac{1}{3}, \pm\dfrac{2}{3}, \pm\dfrac{4}{3} \). Twelve in all. \( \pm 1, \pm 2, \pm 4, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{4}{3} \)

7. Find all roots of \( x^3 - 6x^2 + 11x - 6 \).
   Show the full solution

   Candidates: \( \pm 1, \pm 2, \pm 3, \pm 6 \). \( P(1) = 1 - 6 + 11 - 6 = 0 \). A root. Synthetic division with 1 on \( 1, -6, 11, -6 \): bring down 1; \( -6 + 1 = -5 \); \( 11 - 5 = 6 \); \( -6 + 6 = 0 \). Quotient \( x^2 - 5x + 6 = (x - 2)(x - 3) \). All roots: 1, 2, 3. Check: \( P(2) = 8 - 24 + 22 - 6 = 0 \). Correct. \( P(3) = 27 - 54 + 33 - 6 = 0 \). Correct. Check the sum: it should be \( -\dfrac{b}{a} = 6 \), and \( 1 + 2 + 3 = 6 \). Correct. \( x = 1, 2, 3 \)

8. Find all roots of \( x^3 - x^2 + 4x - 4 \), including complex ones.
   Show the full solution

   Candidates: \( \pm 1, \pm 2, \pm 4 \). \( P(1) = 1 - 1 + 4 - 4 = 0 \). A root. Synthetic division with 1 on \( 1, -1, 4, -4 \): bring down 1; \( -1 + 1 = 0 \); \( 4 + 0 = 4 \); \( -4 + 4 = 0 \). Quotient \( x^2 + 0x + 4 = x^2 + 4 \). That has no real roots, but over the complex numbers \( x^2 = -4 \) gives \( x = \pm 2i \). All roots: \( 1 \), \( 2i \), \( -2i \). Three roots for a cubic, as the fundamental theorem requires, with the non-real pair conjugate, as a real polynomial requires. Check: \( P(2i) = (2i)^3 - (2i)^2 + 4(2i) - 4 = -8i + 4 + 8i - 4 = 0 \). Correct. Grouping gives the same result faster: \( x^2(x - 1) + 4(x - 1) = (x - 1)(x^2 + 4) \). \( x = 1,\; \pm 2i \)

9. Explain why a polynomial of odd degree with real coefficients must have at least one real root.
   Show the full solution

   Two facts combine. First, the total count. By the fundamental theorem, a polynomial of degree \( n \) has exactly \( n \) complex roots counted with multiplicity. Second, the pairing. By the conjugate pair theorem of lesson 2.6, non-real roots of a real polynomial come in conjugate pairs, so the number of non-real roots is even. If \( n \) is odd, subtracting an even number of non-real roots from an odd total leaves an odd number of real roots. An odd number is at least 1, so a real root exists. The graphical version of the same fact. An odd-degree polynomial has opposite end behaviors, so it takes both large positive and large negative values. Being continuous, it must cross zero in between. Both arguments give the same conclusion by different routes, which is a good sign that the conclusion is structural. Why the even case differs. An even degree allows all roots to be non-real, since an even total can be entirely made of pairs. \( x^2 + 1 \) has roots \( \pm i \) and no real root, and its graph never touches the axis. A consequence worth stating. Every cubic with real coefficients crosses the \( x \)-axis at least once, so every cubic equation has at least one real solution. That is why the cubic was solvable by radicals historically while the general question of complex roots took much longer to settle. An odd total minus an even number of non-real roots leaves an odd number of real roots, which is at least one

10. Find all roots of \( 2x^4 - x^3 - 19x^2 + 9x + 9 \).
    Show the full solution

    List candidates. Divisors of 9: \( 1, 3, 9 \). Divisors of 2: \( 1, 2 \). Candidates: \( \pm 1, \pm 3, \pm 9, \pm\dfrac{1}{2}, \pm\dfrac{3}{2}, \pm\dfrac{9}{2} \). Test. \( P(1) = 2 - 1 - 19 + 9 + 9 = 0 \). A root. Divide out. Synthetic division with 1 on \( 2, -1, -19, 9, 9 \): bring down 2; \( -1 + 2 = 1 \); \( -19 + 1 = -18 \); \( 9 - 18 = -9 \); \( 9 - 9 = 0 \). Quotient: \( 2x^3 + x^2 - 18x - 9 \). Test again on the cubic. \( 2 + 1 - 18 - 9 = -24 \) at \( x = 1 \). No. At \( x = -1 \): \( -2 + 1 + 18 - 9 = 8 \). No. At \( x = 3 \): \( 54 + 9 - 54 - 9 = 0 \). A root. Divide again. Synthetic division with 3 on \( 2, 1, -18, -9 \): bring down 2; \( 1 + 6 = 7 \); \( -18 + 21 = 3 \); \( -9 + 9 = 0 \). Quotient: \( 2x^2 + 7x + 3 \). Finish with the quadratic. Discriminant: \( 49 - 24 = 25 \). \( x = \dfrac{-7 \pm 5}{4} \), giving \( x = -\dfrac{1}{2} \) or \( x = -3 \). All four roots: \( 1 \), \( 3 \), \( -\dfrac{1}{2} \), \( -3 \). Verify each in the original. \( P(3) = 162 - 27 - 171 + 27 + 9 = 0 \). Correct. \( P(-3) = 162 + 27 - 171 - 27 + 9 = 0 \). Correct. \( P\left( -\dfrac{1}{2} \right) = 2\left( \dfrac{1}{16} \right) + \dfrac{1}{8} - 19\left( \dfrac{1}{4} \right) - \dfrac{9}{2} + 9 = \dfrac{1}{8} + \dfrac{1}{8} - \dfrac{19}{4} - \dfrac{9}{2} + 9 = \dfrac{1}{4} - \dfrac{19}{4} - \dfrac{18}{4} + \dfrac{36}{4} = \dfrac{0}{4} = 0 \). Correct. Check the count. Degree 4 with four distinct real roots, so no complex roots and no repeated ones. Consistent with the fundamental theorem. Check the product of the roots. It should be \( \dfrac{a_0}{a_n} = \dfrac{9}{2} \) for even degree. \( (1)(3)\left( -\dfrac{1}{2} \right)(-3) = \dfrac{9}{2} \). Correct. Check the sum. It should be \( -\dfrac{a_3}{a_4} = \dfrac{1}{2} \). \( 1 + 3 - \dfrac{1}{2} - 3 = \dfrac{1}{2} \). Correct. The factored form. \( 2(x - 1)(x - 3)\left( x + \dfrac{1}{2} \right)(x + 3) \), or equivalently \( (x - 1)(x - 3)(2x + 1)(x + 3) \). \( x = 1,\; 3,\; -\frac{1}{2},\; -3 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Give the roots of \( y = (x - 1)(x + 4) \).
   Show the full solution

   1 and \( -4 \)

2. Does the graph cross or touch at a root of multiplicity 2?
   Show the full solution

   Touches

3. Give the degree of \( y = (x - 2)^3(x + 1) \).
   Show the full solution

   \( 3 + 1 \). 4

4. Find the \( y \)-intercept of \( y = (x - 1)(x + 2) \).
   Show the full solution

   \( (-1)(2) \). \( -2 \)

5. At most how many turning points can a degree-5 polynomial have?
   Show the full solution

   4

6. Describe the graph of \( y = -(x - 2)^2(x + 1) \) at each root and at its ends.
   Show the full solution

   Degree 3, and the leading term is \( -x^2 \cdot x = -x^3 \), so the leading coefficient is \( -1 \). Odd degree, negative leading coefficient: rises on the left, falls on the right. Root \( x = 2 \), multiplicity 2: touches and turns back. Root \( x = -1 \), multiplicity 1: crosses. \( y \)-intercept: \( -(4)(1) = -4 \). Sign check: at \( x = -2 \), \( -(16)(-1) = 16 \), positive. At \( x = 0 \), \( -4 \), negative. At \( x = 3 \), \( -(1)(4) = -4 \), negative. Negative on both sides of \( x = 2 \), confirming the touch. Rises from the left, crosses at \( -1 \), touches at 2, falls to the right

7. A quartic has roots at \( -1 \), 0 and 2, with 0 having multiplicity 2, and a positive leading coefficient. Write a possible equation and give its \( y \)-intercept.
   Show the full solution

   Multiplicities must total 4: \( 1 + 2 + 1 = 4 \). Correct. \( y = a\,x^2(x + 1)(x - 2) \) with \( a \gt 0 \). Take \( a = 1 \): \( y = x^2(x + 1)(x - 2) \). \( y \)-intercept: \( f(0) = 0 \), since \( x = 0 \) is a root. The graph passes through the origin and touches there. Check the degree: \( x^2 \cdot x \cdot x = x^4 \). Correct, leading coefficient 1, positive. Sign check: at \( x = 1 \), \( (1)(2)(-1) = -2 \), negative. At \( x = -0.5 \), \( (0.25)(0.5)(-2.5) = -0.3125 \), negative. Negative on both sides of 0, confirming the touch. \( y = x^2(x+1)(x-2) \), \( y \)-intercept 0

8. Sketch-describe \( y = (x + 1)^3(x - 2) \), including what happens at each root.
   Show the full solution

   Degree \( 3 + 1 = 4 \), leading term \( x^3 \cdot x = x^4 \), coefficient positive. Both ends rise. Root \( x = -1 \), multiplicity 3: odd, so it crosses, but with a visible flattening against the axis because the cube is very small nearby. Root \( x = 2 \), multiplicity 1: crosses normally. \( y \)-intercept: \( (1)^3(-2) = -2 \). Sign check: at \( x = -2 \), \( (-1)^3(-4) = 4 \), positive. At \( x = 0 \), \( -2 \), negative. At \( x = 3 \), \( (64)(1) = 64 \), positive. The sign changes at both roots, confirming both are crossings. The shape. Falls from the upper left, crosses at \( -1 \) with a flattening, dips to a minimum, crosses at 2, rises. Two turning points, fewer than the three a quartic allows, because the triple root uses up the wiggle room. Crosses flatly at \( -1 \), crosses at 2, both ends rise

9. Explain why a polynomial of degree \( n \) has at most \( n - 1 \) turning points.
   Show the full solution

   A turning point is where the graph changes from increasing to decreasing or the reverse. Between two consecutive turning points the graph is monotonic, so it can cross any given horizontal level at most once in that stretch. If a polynomial had \( n \) turning points, the graph would have \( n + 1 \) monotonic stretches. Choosing a horizontal line at a suitable height would then cut the graph \( n + 1 \) times, giving \( n + 1 \) solutions to an equation of degree \( n \). That contradicts the fact that a degree-\( n \) polynomial equation has at most \( n \) solutions. The calculus version, for later. Turning points occur where the derivative is zero, and the derivative of a degree-\( n \) polynomial has degree \( n - 1 \), so it has at most \( n - 1 \) roots. That is the same bound reached by a shorter route, and it is why this lesson can state the fact without proving it from calculus. When the maximum is not reached. \( y = x^4 \) has degree 4 but only one turning point, and \( y = x^3 \) has degree 3 and none at all. Repeated roots and flattening consume the available turns, which is why the statement is "at most" rather than "exactly". A useful check when sketching. Counting the turns in a sketch and comparing with \( n - 1 \) catches a graph drawn with too many wiggles. A cubic cannot have three humps. More than \( n-1 \) turns would let a horizontal line cross more than \( n \) times, giving too many solutions to a degree-\( n \) equation

10. A quartic with positive leading coefficient has \( f(-2) = 0 \), \( f(1) = 0 \), \( f(4) = 0 \), touches the axis at \( x = 1 \), and has \( f(0) = 16 \). Find it and describe its graph.
    Show the full solution

    Assign multiplicities. Touching at \( x = 1 \) means even multiplicity there, and the degree is 4 with three distinct roots, so the multiplicities must be \( 1 + 2 + 1 = 4 \). The double root is at 1. Write the form. \( f(x) = a(x + 2)(x - 1)^2(x - 4) \). Find \( a \) from the \( y \)-intercept. \( f(0) = a(2)(1)(-4) = -8a \). Setting \( -8a = 16 \) gives \( a = -2 \). A problem. The question specified a positive leading coefficient, and the leading coefficient here is \( a = -2 \), which is negative. The four stated conditions are therefore inconsistent with each other. Check that carefully before concluding. The leading term is \( a \cdot x \cdot x^2 \cdot x = a x^4 \), so the leading coefficient is exactly \( a \). And \( f(0) \) with \( a \) positive would be \( -8a \), which is negative, never 16. So no quartic satisfies all five conditions. Why this happens. With both ends rising and roots at \( -2 \), 1 and 4, the graph is above the axis outside \( [-2, 4] \) and its sign inside is determined. Just to the right of \( -2 \) it must be below the axis, and since the double root at 1 does not change the sign, it stays below all the way to 4. The point \( x = 0 \) lies in that interval, so \( f(0) \) must be negative. A value of \( +16 \) is impossible. Resolve it the useful way. Drop the leading-coefficient condition, which is the one the other four contradict: \( f(x) = -2(x + 2)(x - 1)^2(x - 4) \). Verify. \( f(0) = -2(2)(1)(-4) = 16 \). Correct. \( f(-2) = 0 \), \( f(1) = 0 \), \( f(4) = 0 \). Correct. Describe the graph. Leading coefficient \( -2 \), even degree, so both ends fall. Crosses at \( -2 \), touches at 1, crosses at 4. Sign check: at \( x = -3 \), \( -2(-1)(16)(-7) = -224 \), negative. At \( x = 0 \), \( +16 \), positive. At \( x = 2 \), \( -2(4)(1)(-2) = 16 \), positive. At \( x = 5 \), \( -2(7)(16)(1) = -224 \), negative. Positive on both sides of 1, confirming the touch. What the exercise demonstrates. Conditions on a polynomial can over-determine it, and the right response is to notice the contradiction and say so rather than to force an answer. Here the sign of the \( y \)-intercept was already fixed by the roots and the end behavior, so specifying it independently was one condition too many. No quartic meets all five conditions; dropping the positive leading coefficient gives \( f(x) = -2(x+2)(x-1)^2(x-4) \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. State excluded values every time.

1. Simplify \( \dfrac{6x^2}{2x} \).
   Show the full solution

   \( 3x \), with \( x \ne 0 \)

2. Simplify \( \dfrac{x^2 - 4}{x - 2} \).
   Show the full solution

   \( \dfrac{(x-2)(x+2)}{x-2} \). \( x + 2 \), with \( x \ne 2 \)

3. Can \( \dfrac{x + 5}{5} \) be simplified?
   Show the full solution

   The 5 on top is a term, not a factor. No

4. Simplify \( \dfrac{x - 4}{4 - x} \).
   Show the full solution

   Opposites. \( -1 \), with \( x \ne 4 \)

5. State the excluded values of \( \dfrac{1}{x^2 - 16} \).
   Show the full solution

   \( x \ne 4 \) and \( x \ne -4 \)

6. Simplify \( \dfrac{x^2 + 5x + 6}{x^2 - x - 6} \).
   Show the full solution

   Numerator: \( (x + 2)(x + 3) \). Denominator: \( (x - 3)(x + 2) \). Excluded: \( x = 3 \) and \( x = -2 \). Cancel \( (x + 2) \): \( \dfrac{x + 3}{x - 3} \). Check at \( x = 0 \): the original gives \( \dfrac{6}{-6} = -1 \), and the simplified form gives \( \dfrac{3}{-3} = -1 \). Agrees. \( \frac{x+3}{x-3} \), with \( x \ne 3, -2 \)

7. Simplify \( \dfrac{3x^2 - 12x}{x^2 - 16} \).
   Show the full solution

   Numerator: \( 3x(x - 4) \), taking the common factor first. Denominator: \( (x - 4)(x + 4) \). Excluded: \( x = 4 \) and \( x = -4 \). Cancel \( (x - 4) \): \( \dfrac{3x}{x + 4} \). Check at \( x = 1 \): the original gives \( \dfrac{3 - 12}{1 - 16} = \dfrac{-9}{-15} = \dfrac{3}{5} \), and the simplified form gives \( \dfrac{3}{5} \). Agrees. \( \frac{3x}{x+4} \), with \( x \ne \pm 4 \)

8. Simplify \( \dfrac{x^3 - 8}{x^2 - 4} \).
   Show the full solution

   Numerator is a difference of cubes: \( x^3 - 8 = (x - 2)(x^2 + 2x + 4) \). Denominator: \( (x - 2)(x + 2) \). Excluded: \( x = 2 \) and \( x = -2 \). Cancel \( (x - 2) \): \( \dfrac{x^2 + 2x + 4}{x + 2} \). Check at \( x = 0 \): the original gives \( \dfrac{-8}{-4} = 2 \), and the simplified form gives \( \dfrac{4}{2} = 2 \). Agrees. The quadratic factor has discriminant \( 4 - 16 = -12 \), so nothing further cancels. \( \frac{x^2+2x+4}{x+2} \), with \( x \ne \pm 2 \)

9. Explain why a term cannot be canceled, using a numerical counterexample.
   Show the full solution

   Canceling is division, and division distributes over a product, not over a sum. \( \dfrac{ab}{b} = a \) because the \( b \) divides the whole numerator. But \( \dfrac{a + b}{b} \) is \( \dfrac{a}{b} + 1 \), not \( a \), because only part of the numerator has a \( b \) in it. The counterexample. Take \( \dfrac{6 + 4}{4} \). The true value is \( \dfrac{10}{4} = 2.5 \). Canceling the 4s would give 6, which is wrong by a factor of more than two. The same test on an expression. Take \( \dfrac{x + 3}{3} \) at \( x = 6 \): the true value is \( \dfrac{9}{3} = 3 \), while the false cancellation gives 6. Different. Why the error is tempting. The 3 appears above and below, and the eye pattern-matches on appearance rather than on structure. The defense is to factor first: if a factored numerator does not contain the denominator as a whole factor, nothing cancels. Where this recurs. The same mistake appears as the linearity error throughout the course: \( \sqrt{a + b} \ne \sqrt{a} + \sqrt{b} \), \( \log(a + b) \ne \log a + \log b \). In every case an operation is being distributed across a sum when it distributes only across a product. Division distributes over products, not sums; \( \frac{6+4}{4} = 2.5 \), not 6

10. Simplify \( \dfrac{2x^2 + 5x - 3}{6 - x - x^2} \) and state the excluded values.
    Show the full solution

    Factor the numerator. Look for two numbers multiplying to \( 2 \times (-3) = -6 \) and adding to 5: they are 6 and \( -1 \). \( 2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (x + 3)(2x - 1) \). Check: \( (x+3)(2x-1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3 \). Correct. Factor the denominator. It is written in ascending order and has a negative leading term, so rearrange and factor out \( -1 \) first: \( 6 - x - x^2 = -(x^2 + x - 6) = -(x + 3)(x - 2) \). Check: \( -(x^2 + x - 6) = -x^2 - x + 6 \). Correct. State the exclusions from the original denominator. It is zero when \( x = -3 \) or \( x = 2 \). Cancel. \( \dfrac{(x + 3)(2x - 1)}{-(x + 3)(x - 2)} = \dfrac{2x - 1}{-(x - 2)} = \dfrac{2x - 1}{2 - x} \). The negative can be placed in the denominator to turn \( -(x - 2) \) into \( 2 - x \), or kept out front as \( -\dfrac{2x - 1}{x - 2} \). Both are correct. Check at \( x = 0 \). Original: \( \dfrac{-3}{6} = -\dfrac{1}{2} \). Simplified: \( \dfrac{-1}{2} = -\dfrac{1}{2} \). Agrees. Check at \( x = 1 \). Original: \( \dfrac{2 + 5 - 3}{6 - 1 - 1} = \dfrac{4}{4} = 1 \). Simplified: \( \dfrac{2 - 1}{2 - 1} = 1 \). Agrees. Note which exclusion is invisible afterward. The simplified form shows only \( x \ne 2 \). The exclusion \( x \ne -3 \) came from a canceled factor and must be carried along explicitly. At \( x = -3 \) the graph has a hole, which lesson 4.6 locates at \( \left( -3,\, \dfrac{-7}{5} \right) \), since the simplified form there gives \( \dfrac{-6 - 1}{2 + 3} = -\dfrac{7}{5} \). \( \frac{2x-1}{2-x} \), with \( x \ne -3 \) and \( x \ne 2 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( \dfrac{2}{x} \cdot \dfrac{x}{3} \).
   Show the full solution

   \( \frac{2}{3} \), with \( x \ne 0 \)

2. Simplify \( \dfrac{x}{4} \div \dfrac{x}{2} \).
   Show the full solution

   \( \dfrac{x}{4} \cdot \dfrac{2}{x} = \dfrac{2}{4} \). \( \frac{1}{2} \), with \( x \ne 0 \)

3. Which fraction gets inverted in a division?
   Show the full solution

   The second one, the divisor

4. Simplify \( \dfrac{x + 1}{x} \cdot \dfrac{x}{x + 1} \).
   Show the full solution

   Everything cancels. 1, with \( x \ne 0, -1 \)

5. Is a common denominator needed to multiply?
   Show the full solution

   No

6. Simplify \( \dfrac{x^2 - 9}{x + 1} \cdot \dfrac{x^2 - 1}{x - 3} \).
   Show the full solution

   Factor: \( \dfrac{(x-3)(x+3)}{x+1} \cdot \dfrac{(x-1)(x+1)}{x-3} \). Exclusions: \( x \ne -1 \) and \( x \ne 3 \). Cancel \( (x - 3) \) and \( (x + 1) \) across. Left: \( (x + 3)(x - 1) \). Check at \( x = 0 \): the original gives \( \dfrac{-9}{1} \cdot \dfrac{-1}{-3} = -9 \times \dfrac{1}{3} = -3 \), and \( (3)(-1) = -3 \). Agrees. \( (x+3)(x-1) \), with \( x \ne -1, 3 \)

7. Simplify \( \dfrac{x^2 + 2x}{x^2 - 9} \div \dfrac{x + 2}{x - 3} \).
   Show the full solution

   Invert the divisor: \( \dfrac{x^2 + 2x}{x^2 - 9} \cdot \dfrac{x - 3}{x + 2} \). Factor: \( \dfrac{x(x + 2)}{(x-3)(x+3)} \cdot \dfrac{x - 3}{x + 2} \). Exclusions: \( x \ne 3 \) and \( x \ne -3 \) from the first denominator, \( x \ne 3 \) from the divisor's denominator, and \( x \ne -2 \) from the divisor's numerator. Together: \( x \ne 3, -3, -2 \). Cancel \( (x + 2) \) and \( (x - 3) \). Left: \( \dfrac{x}{x + 3} \). Check at \( x = 0 \): the original gives \( \dfrac{0}{-9} \div \dfrac{2}{-3} = 0 \div \left( -\dfrac{2}{3} \right) = 0 \), and \( \dfrac{0}{3} = 0 \). Agrees. \( \frac{x}{x+3} \), with \( x \ne 3, -3, -2 \)

8. Simplify \( \dfrac{2x^2 - 2}{x^2 + 4x + 3} \cdot \dfrac{x^2 + 6x + 9}{4x - 4} \).
   Show the full solution

   Factor each part. \( 2x^2 - 2 = 2(x - 1)(x + 1) \). \( x^2 + 4x + 3 = (x + 1)(x + 3) \). \( x^2 + 6x + 9 = (x + 3)^2 \). \( 4x - 4 = 4(x - 1) \). Exclusions: \( x \ne -1 \), \( x \ne -3 \), \( x \ne 1 \). Assemble: \( \dfrac{2(x-1)(x+1)}{(x+1)(x+3)} \cdot \dfrac{(x+3)^2}{4(x-1)} \). Cancel \( (x - 1) \), \( (x + 1) \), and one \( (x + 3) \). Numbers: \( \dfrac{2}{4} = \dfrac{1}{2} \). Left: \( \dfrac{x + 3}{2} \). Check at \( x = 0 \): the original gives \( \dfrac{-2}{3} \cdot \dfrac{9}{-4} = \dfrac{-18}{-12} = \dfrac{3}{2} \), and \( \dfrac{0 + 3}{2} = \dfrac{3}{2} \). Agrees. \( \frac{x+3}{2} \), with \( x \ne 1, -1, -3 \)

9. Explain why the answer to a multiplication can be a polynomial even though both inputs were fractions.
   Show the full solution

   A rational expression is a fraction of polynomials, and a polynomial is the special case where the denominator is 1. So polynomials are already rational expressions, and nothing unusual has happened when one appears as an answer. How it happens mechanically. Canceling removes factors from the denominator. If every denominator factor finds a partner in some numerator, the denominator reduces to 1 and the result is a polynomial. The numerical analogy. \( \dfrac{3}{4} \times \dfrac{8}{3} = 2 \), an integer, although neither factor was. Multiplying fractions can land on a whole number whenever the denominators divide out, and the polynomial case is identical. The one thing that does not go away. The exclusions remain. In problem 6 the answer \( (x + 3)(x - 1) \) is a polynomial, defined for every real number, but the expression it came from is undefined at \( x = -1 \) and \( x = 3 \). The answer is that polynomial with two points removed, and the graph has holes there. Denominators can cancel completely, leaving a denominator of 1; the excluded values still apply

10. Simplify \( \dfrac{x^3 - 27}{x^2 - 4} \div \dfrac{x^2 + 3x + 9}{x^2 + 2x} \), stating all excluded values.
    Show the full solution

    Invert the divisor. \( \dfrac{x^3 - 27}{x^2 - 4} \cdot \dfrac{x^2 + 2x}{x^2 + 3x + 9} \). Factor every part. \( x^3 - 27 = (x - 3)(x^2 + 3x + 9) \), a difference of cubes. \( x^2 - 4 = (x - 2)(x + 2) \). \( x^2 + 2x = x(x + 2) \). \( x^2 + 3x + 9 \) has discriminant \( 9 - 36 = -27 \), so it is irreducible over the reals and stays as it is. Collect the exclusions before canceling. From the first denominator \( x^2 - 4 \): \( x \ne 2 \) and \( x \ne -2 \). From the divisor's denominator \( x^2 + 2x = x(x + 2) \): \( x \ne 0 \) and \( x \ne -2 \). From the divisor's numerator \( x^2 + 3x + 9 \), which becomes a denominator: it is never zero for real \( x \), since its discriminant is negative, so it excludes nothing. Together: \( x \ne 2 \), \( x \ne -2 \), \( x \ne 0 \). Assemble and cancel. \( \dfrac{(x - 3)(x^2 + 3x + 9)}{(x - 2)(x + 2)} \cdot \dfrac{x(x + 2)}{x^2 + 3x + 9} \). The irreducible quadratic cancels across, and so does \( (x + 2) \). Left: \( \dfrac{x(x - 3)}{x - 2} \). Check at \( x = 1 \). Original: \( \dfrac{1 - 27}{1 - 4} \div \dfrac{1 + 3 + 9}{1 + 2} = \dfrac{-26}{-3} \div \dfrac{13}{3} = \dfrac{26}{3} \times \dfrac{3}{13} = 2 \). Answer: \( \dfrac{1(1 - 3)}{1 - 2} = \dfrac{-2}{-1} = 2 \). Agrees. Check at \( x = 4 \). Original: \( \dfrac{64 - 27}{16 - 4} \div \dfrac{16 + 12 + 9}{16 + 8} = \dfrac{37}{12} \div \dfrac{37}{24} = \dfrac{37}{12} \times \dfrac{24}{37} = 2 \). Answer: \( \dfrac{4(1)}{2} = 2 \). Agrees. Why the irreducible quadratic mattered. Spotting that \( x^3 - 27 \) contains \( x^2 + 3x + 9 \) as a factor is what makes the problem collapse. Without recognizing the difference of cubes, the expression looks unsimplifiable. The three special factorizations of lesson 3.5 earn their place here. \( \frac{x(x-3)}{x-2} \), with \( x \ne 0, 2, -2 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( \dfrac{2}{x} + \dfrac{3}{x} \).
   Show the full solution

   \( \frac{5}{x} \), with \( x \ne 0 \)

2. Simplify \( \dfrac{7}{x} - \dfrac{4}{x} \).
   Show the full solution

   \( \frac{3}{x} \), with \( x \ne 0 \)

3. Give the least common denominator of \( \dfrac{1}{x} \) and \( \dfrac{1}{x + 1} \).
   Show the full solution

   \( x(x+1) \)

4. Give the least common denominator of \( \dfrac{1}{x^2} \) and \( \dfrac{1}{x^3} \).
   Show the full solution

   Highest power of the only factor. \( x^3 \)

5. Simplify \( \dfrac{1}{2} + \dfrac{1}{x} \).
   Show the full solution

   Common denominator \( 2x \): \( \dfrac{x + 2}{2x} \). \( \frac{x+2}{2x} \), with \( x \ne 0 \)

6. Simplify \( \dfrac{4}{x - 2} - \dfrac{3}{x + 1} \).
   Show the full solution

   Least common denominator \( (x-2)(x+1) \). Exclusions: \( x \ne 2, -1 \). \( \dfrac{4(x + 1) - 3(x - 2)}{(x-2)(x+1)} \). Numerator: \( 4x + 4 - 3x + 6 = x + 10 \). Note \( -3(x - 2) = -3x + 6 \), with both signs changing. \( \dfrac{x + 10}{(x-2)(x+1)} \). Check at \( x = 0 \): \( \dfrac{4}{-2} - \dfrac{3}{1} = -2 - 3 = -5 \), and \( \dfrac{10}{(-2)(1)} = -5 \). Agrees. \( \frac{x+10}{(x-2)(x+1)} \), with \( x \ne 2, -1 \)

7. Simplify \( \dfrac{3}{x^2 - x} + \dfrac{2}{x} \).
   Show the full solution

   Factor: \( x^2 - x = x(x - 1) \). Least common denominator: \( x(x - 1) \). Exclusions: \( x \ne 0, 1 \). Second fraction needs \( (x - 1) \): \( \dfrac{3 + 2(x - 1)}{x(x - 1)} = \dfrac{3 + 2x - 2}{x(x-1)} = \dfrac{2x + 1}{x(x-1)} \). Check at \( x = 2 \): \( \dfrac{3}{2} + \dfrac{2}{2} = 1.5 + 1 = 2.5 \), and \( \dfrac{5}{2(1)} = 2.5 \). Agrees. \( \frac{2x+1}{x(x-1)} \), with \( x \ne 0, 1 \)

8. Simplify \( \dfrac{x}{x^2 - 4} - \dfrac{1}{x - 2} \).
   Show the full solution

   Factor: \( x^2 - 4 = (x-2)(x+2) \), which already contains \( x - 2 \). Least common denominator: \( (x-2)(x+2) \). Exclusions: \( x \ne \pm 2 \). \( \dfrac{x - (x + 2)}{(x-2)(x+2)} \). Numerator: \( x - x - 2 = -2 \). Both terms of \( (x + 2) \) received the minus. \( \dfrac{-2}{(x-2)(x+2)} \), or \( \dfrac{-2}{x^2 - 4} \). Check at \( x = 0 \): \( \dfrac{0}{-4} - \dfrac{1}{-2} = 0 + 0.5 = 0.5 \), and \( \dfrac{-2}{-4} = 0.5 \). Agrees. Check at \( x = 3 \): \( \dfrac{3}{5} - \dfrac{1}{1} = -0.4 \), and \( \dfrac{-2}{5} = -0.4 \). Agrees. \( \frac{-2}{(x-2)(x+2)} \), with \( x \ne \pm 2 \)

9. Explain why the least common denominator is built from factorizations rather than by multiplying the denominators.
   Show the full solution

   Multiplying the denominators always produces a common denominator, since each original divides the product. It is just rarely the smallest one, and using an unnecessarily large denominator makes every subsequent step harder. A concrete comparison. For \( \dfrac{5}{x^2 - 9} - \dfrac{2}{x + 3} \), multiplying gives \( (x^2 - 9)(x + 3) = (x - 3)(x + 3)^2 \), a cubic. The numerators would then be larger, and the final answer would contain a common factor of \( (x + 3) \) that has to be canceled at the end. Building from factorizations gives \( (x - 3)(x + 3) \) directly, and the answer comes out already in lowest terms. Why factoring is unavoidable anyway. Even with the larger denominator, the factorizations are needed at the end to cancel. Doing them first saves the work rather than adding to it. The case where it matters most. When the denominators share a repeated factor, such as \( (x-1)^2 \) and \( (x-1)^3 \), multiplying gives \( (x-1)^5 \) where \( (x-1)^3 \) suffices. The difference grows with the exponents, and with three or more fractions it becomes substantial. Multiplying gives a valid but unnecessarily large denominator, forcing bigger numerators and a cancellation at the end

10. Simplify \( \dfrac{2}{x^2 - x - 6} + \dfrac{3}{x^2 - 9} - \dfrac{1}{x^2 + 5x + 6} \).
    Show the full solution

    Factor all three denominators. \( x^2 - x - 6 = (x - 3)(x + 2) \). \( x^2 - 9 = (x - 3)(x + 3) \). \( x^2 + 5x + 6 = (x + 2)(x + 3) \). Build the least common denominator. The distinct factors are \( (x - 3) \), \( (x + 2) \) and \( (x + 3) \), each appearing to the first power only: \( (x - 3)(x + 2)(x + 3) \). Exclusions. \( x \ne 3 \), \( x \ne -2 \), \( x \ne -3 \). Rewrite each fraction with the missing factor. First needs \( (x + 3) \): \( \dfrac{2(x + 3)}{(x-3)(x+2)(x+3)} \). Second needs \( (x + 2) \): \( \dfrac{3(x + 2)}{(x-3)(x+2)(x+3)} \). Third needs \( (x - 3) \): \( \dfrac{1(x - 3)}{(x-3)(x+2)(x+3)} \). Combine the numerators, bracketing the subtracted one. \( 2(x + 3) + 3(x + 2) - (x - 3) \) \( = 2x + 6 + 3x + 6 - x + 3 \) \( = 4x + 15 \). The \( -(x - 3) \) contributed \( -x + 3 \), with both signs flipped. The answer. \( \dfrac{4x + 15}{(x - 3)(x + 2)(x + 3)} \), with \( x \ne 3, -2, -3 \). Check whether anything cancels. The numerator is zero at \( x = -\dfrac{15}{4} \), which is none of the denominator's roots, so nothing cancels and the answer is in lowest terms. Check at \( x = 0 \). Original: \( \dfrac{2}{-6} + \dfrac{3}{-9} - \dfrac{1}{6} = -\dfrac{1}{3} - \dfrac{1}{3} - \dfrac{1}{6} = -\dfrac{5}{6} \). Answer: \( \dfrac{15}{(-3)(2)(3)} = \dfrac{15}{-18} = -\dfrac{5}{6} \). Agrees. Check at \( x = 1 \). Original: \( \dfrac{2}{-6} + \dfrac{3}{-8} - \dfrac{1}{12} = -\dfrac{1}{3} - \dfrac{3}{8} - \dfrac{1}{12} \). Over 24: \( -\dfrac{8}{24} - \dfrac{9}{24} - \dfrac{2}{24} = -\dfrac{19}{24} \). Answer: \( \dfrac{19}{(-2)(3)(4)} = \dfrac{19}{-24} = -\dfrac{19}{24} \). Agrees. What made this manageable. All three denominators were built from the same three factors, so the least common denominator had only three factors rather than six. Multiplying the denominators together would have produced a degree-6 expression with every factor duplicated, and the final answer would have needed a substantial cancellation. Factoring first is what kept the work small. \( \frac{4x+15}{(x-3)(x+2)(x+3)} \), with \( x \ne 3, -2, -3 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( \dfrac{\frac{1}{2}}{\frac{3}{4}} \).
   Show the full solution

   \( \dfrac{1}{2} \cdot \dfrac{4}{3} \). \( \frac{2}{3} \)

2. Simplify \( \dfrac{\frac{x}{3}}{\frac{x}{6}} \).
   Show the full solution

   \( \dfrac{x}{3} \cdot \dfrac{6}{x} \). 2, with \( x \ne 0 \)

3. What is the fastest multiplier for \( \dfrac{\frac{1}{x}}{\frac{1}{x^2}} \)?
   Show the full solution

   \( x^2 \)

4. Simplify that expression.
   Show the full solution

   Top becomes \( x \), bottom becomes 1. \( x \), with \( x \ne 0 \)

5. Why is multiplying top and bottom by the same thing allowed?
   Show the full solution

   It is multiplication by 1

6. Simplify \( \dfrac{1 + \frac{1}{x}}{1 - \frac{1}{x}} \).
   Show the full solution

   Multiply top and bottom by \( x \): Numerator: \( x + 1 \). Denominator: \( x - 1 \). \( \dfrac{x + 1}{x - 1} \). Exclusions: \( x \ne 0 \) from the small fractions, and \( x \ne 1 \) because the main denominator \( 1 - \dfrac{1}{x} \) is zero there. Check at \( x = 3 \): \( \dfrac{1 + \frac{1}{3}}{1 - \frac{1}{3}} = \dfrac{4/3}{2/3} = 2 \), and \( \dfrac{4}{2} = 2 \). Agrees. \( \frac{x+1}{x-1} \), with \( x \ne 0, 1 \)

7. Simplify \( \dfrac{\frac{2}{x} - \frac{2}{3}}{x - 3} \).
   Show the full solution

   Multiply top and bottom by \( 3x \), the least common denominator of the small fractions: Numerator: \( 3x\left( \dfrac{2}{x} - \dfrac{2}{3} \right) = 6 - 2x \). Denominator: \( 3x(x - 3) \). \( \dfrac{6 - 2x}{3x(x - 3)} = \dfrac{-2(x - 3)}{3x(x - 3)} = \dfrac{-2}{3x} \). Factoring out \( -2 \) revealed the matching factor. Exclusions: \( x \ne 0 \) and \( x \ne 3 \). Check at \( x = 1 \): \( \dfrac{2 - \frac{2}{3}}{-2} = \dfrac{4/3}{-2} = -\dfrac{2}{3} \), and \( \dfrac{-2}{3} = -\dfrac{2}{3} \). Agrees. \( \frac{-2}{3x} \), with \( x \ne 0, 3 \)

8. Simplify \( \dfrac{\frac{1}{x + 2} + \frac{1}{x}}{\frac{1}{x}} \).
   Show the full solution

   Multiply top and bottom by \( x(x + 2) \): Numerator: \( x(x+2)\left( \dfrac{1}{x+2} + \dfrac{1}{x} \right) = x + (x + 2) = 2x + 2 \). Denominator: \( x(x+2) \cdot \dfrac{1}{x} = x + 2 \). \( \dfrac{2x + 2}{x + 2} = \dfrac{2(x + 1)}{x + 2} \). Exclusions: \( x \ne 0 \) and \( x \ne -2 \). The main denominator \( \dfrac{1}{x} \) is never zero, so it adds nothing further. Check at \( x = 2 \): \( \dfrac{\frac{1}{4} + \frac{1}{2}}{\frac{1}{2}} = \dfrac{0.75}{0.5} = 1.5 \), and \( \dfrac{2(3)}{4} = 1.5 \). Agrees. \( \frac{2(x+1)}{x+2} \), with \( x \ne 0, -2 \)

9. Explain when method one is preferable to method two.
   Show the full solution

   Method two, multiplying through by the least common denominator, is faster whenever the top or the bottom is a sum of several fractions, because it clears them all at once instead of combining them first. Method one, combining and then inverting, is preferable when the numerator and denominator are each already a single fraction. Then there is nothing to combine, and the problem is just a division. A case for method one. \( \dfrac{\frac{x^2 - 1}{x + 3}}{\frac{x - 1}{x^2 - 9}} \) is immediately \( \dfrac{x^2 - 1}{x + 3} \cdot \dfrac{x^2 - 9}{x - 1} \), which factors and cancels to \( (x + 1)(x - 3) \). Multiplying through by a common denominator here would be needless work. A case for method two. \( \dfrac{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}{\frac{1}{xy}} \) has three fractions on top. Combining them takes three steps; multiplying through by \( xyz \) takes one. The practical rule. Count the fractions. One over one means method one. Several on either side means method two. Method one is better when the top and bottom are each already a single fraction; method two when either is a sum of several

10. Simplify \( \dfrac{\frac{1}{x^2} - \frac{1}{9}}{\frac{1}{x} + \frac{1}{3}} \), state all exclusions, and verify at two values.
    Show the full solution

    Choose the multiplier. The small denominators are \( x^2 \), 9, \( x \) and 3. The least common denominator is \( 9x^2 \). Multiply the numerator. \( 9x^2\left( \dfrac{1}{x^2} - \dfrac{1}{9} \right) = 9 - x^2 \). Multiply the denominator. \( 9x^2\left( \dfrac{1}{x} + \dfrac{1}{3} \right) = 9x + 3x^2 \). Factor both. Numerator: \( 9 - x^2 = (3 - x)(3 + x) \). Denominator: \( 9x + 3x^2 = 3x(3 + x) \). Cancel. \( \dfrac{(3 - x)(3 + x)}{3x(3 + x)} = \dfrac{3 - x}{3x} \). Collect every exclusion. From the small fractions: \( x \ne 0 \). From the main denominator: \( \dfrac{1}{x} + \dfrac{1}{3} = 0 \) gives \( \dfrac{1}{x} = -\dfrac{1}{3} \), so \( x = -3 \). Excluded. Together: \( x \ne 0 \) and \( x \ne -3 \). Note that \( x = 3 \) is not excluded: the numerator vanishes there, giving an answer of 0, which is perfectly legal. Verify at \( x = 1 \). Original: \( \dfrac{1 - \frac{1}{9}}{1 + \frac{1}{3}} = \dfrac{8/9}{4/3} = \dfrac{8}{9} \times \dfrac{3}{4} = \dfrac{2}{3} \). Answer: \( \dfrac{3 - 1}{3} = \dfrac{2}{3} \). Agrees. Verify at \( x = 6 \). Original: \( \dfrac{\frac{1}{36} - \frac{1}{9}}{\frac{1}{6} + \frac{1}{3}} \). Numerator: \( \dfrac{1}{36} - \dfrac{4}{36} = -\dfrac{3}{36} = -\dfrac{1}{12} \). Denominator: \( \dfrac{1}{6} + \dfrac{2}{6} = \dfrac{1}{2} \). Quotient: \( -\dfrac{1}{12} \times 2 = -\dfrac{1}{6} \). Answer: \( \dfrac{3 - 6}{18} = \dfrac{-3}{18} = -\dfrac{1}{6} \). Agrees. Why this one rewards recognizing the structure. The numerator is a difference of squares in \( \dfrac{1}{x} \) and \( \dfrac{1}{3} \), and the denominator is their sum. So the whole expression is \( \dfrac{a^2 - b^2}{a + b} = a - b \) with \( a = \dfrac{1}{x} \) and \( b = \dfrac{1}{3} \), giving \( \dfrac{1}{x} - \dfrac{1}{3} = \dfrac{3 - x}{3x} \) directly. Same answer, no common denominator needed. Spotting the algebraic identity inside a complex fraction is worth a moment's look before starting the mechanical route. \( \frac{3-x}{3x} \), with \( x \ne 0 \) and \( x \ne -3 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Check every answer.

1. Solve \( \dfrac{x}{3} = 4 \).
   Show the full solution

   \( x = 12 \)

2. Solve \( \dfrac{6}{x} = 2 \).
   Show the full solution

   \( 6 = 2x \). \( x = 3 \)

3. What value must be excluded when solving \( \dfrac{1}{x - 5} = 2 \)?
   Show the full solution

   \( x = 5 \)

4. Solve that equation.
   Show the full solution

   \( 1 = 2(x - 5) \), so \( 1 = 2x - 10 \) and \( x = 5.5 \). Not excluded. Check: \( \dfrac{1}{0.5} = 2 \). Correct. \( x = 5.5 \)

5. Why might a solution have to be discarded?
   Show the full solution

   It may make a denominator zero

6. Solve \( \dfrac{3}{x + 1} + 1 = \dfrac{5}{x + 1} \).
   Show the full solution

   Least common denominator \( x + 1 \), excluding \( x \ne -1 \). Multiply every term: \( 3 + (x + 1) = 5 \). \( x + 4 = 5 \), so \( x = 1 \). Not excluded. Check: \( \dfrac{3}{2} + 1 = 2.5 \), and \( \dfrac{5}{2} = 2.5 \). Correct. \( x = 1 \)

7. Solve \( \dfrac{2x}{x - 3} = \dfrac{6}{x - 3} + 4 \).
   Show the full solution

   Least common denominator \( x - 3 \), excluding \( x \ne 3 \). Multiply every term, including the 4: \( 2x = 6 + 4(x - 3) \). \( 2x = 6 + 4x - 12 = 4x - 6 \). \( -2x = -6 \), so \( x = 3 \). That is the excluded value, so it is extraneous. Substituting confirms: \( \dfrac{6}{0} \) is undefined. No solution

8. Solve \( \dfrac{1}{x - 1} + \dfrac{2}{x + 1} = \dfrac{4}{x^2 - 1} \).
   Show the full solution

   Factor: \( x^2 - 1 = (x - 1)(x + 1) \), which is the least common denominator. Exclusions: \( x \ne 1 \) and \( x \ne -1 \). Multiply every term: \( (x + 1) + 2(x - 1) = 4 \). \( x + 1 + 2x - 2 = 4 \), so \( 3x - 1 = 4 \) and \( x = \dfrac{5}{3} \). Not excluded. Check: \( \dfrac{1}{2/3} + \dfrac{2}{8/3} = \dfrac{3}{2} + \dfrac{6}{8} = 1.5 + 0.75 = 2.25 \). Right side: \( x^2 - 1 = \dfrac{25}{9} - 1 = \dfrac{16}{9} \), so \( \dfrac{4}{16/9} = \dfrac{36}{16} = 2.25 \). Correct. \( x = \frac{5}{3} \)

9. Explain why squaring and clearing denominators can add solutions but never lose them.
   Show the full solution

   Both operations transform an equation into a consequence of it. If \( A = B \) then \( A^2 = B^2 \), and if \( A = B \) then \( kA = kB \) for any \( k \). So anything satisfying the original also satisfies the transformed equation. That is exactly the statement that no solution is lost: every original solution survives into the new solution set. Why the reverse can fail. The implications do not run backward. \( A^2 = B^2 \) allows \( A = -B \) as well as \( A = B \), so squaring admits sign flips. And \( kA = kB \) is automatic when \( k = 0 \), whatever \( A \) and \( B \) are, so multiplying by a variable quantity admits the values that zero it. The consequence for practice. The transformed equation's solution set contains the original's and may be strictly larger. So the candidates found are a superset of the true solutions, and checking is a filtering step rather than a verification of arithmetic. Why this is reassuring. Since nothing is lost, there is no risk of missing a solution by clearing denominators. The only risk is including too many, and that is entirely fixable by checking. An operation that could lose solutions, such as dividing both sides by a variable expression, is genuinely dangerous and should be avoided: dividing \( x^2 = x \) by \( x \) loses the solution \( x = 0 \). Both operations preserve every original solution but are not reversible, so the new solution set can be larger

10. Solve \( \dfrac{x}{x - 2} - \dfrac{2}{x + 2} = \dfrac{8}{x^2 - 4} \) completely, checking every candidate.
    Show the full solution

    Factor and find the least common denominator. \( x^2 - 4 = (x - 2)(x + 2) \), which contains both other denominators. Least common denominator: \( (x - 2)(x + 2) \). Exclusions. \( x \ne 2 \) and \( x \ne -2 \). Multiply every term. First term: \( \dfrac{x}{x-2} \cdot (x-2)(x+2) = x(x + 2) \). Second term: \( \dfrac{2}{x+2} \cdot (x-2)(x+2) = 2(x - 2) \). Right side: \( \dfrac{8}{(x-2)(x+2)} \cdot (x-2)(x+2) = 8 \). Equation: \( x(x + 2) - 2(x - 2) = 8 \). Expand and solve. \( x^2 + 2x - 2x + 4 = 8 \). \( x^2 + 4 = 8 \), so \( x^2 = 4 \) and \( x = \pm 2 \). Apply the exclusion check. \( x = 2 \) is excluded. Extraneous. \( x = -2 \) is excluded. Extraneous. Both candidates fail, so the equation has no solution. Confirm by substitution. At \( x = 2 \) the first term is \( \dfrac{2}{0} \), undefined. At \( x = -2 \) the second term is \( \dfrac{2}{0} \), undefined. Neither value is even in the domain of the equation. Why this happened. Clearing the denominators produced \( x^2 + 4 = 8 \), whose solutions are precisely the two values that were forbidden. The multiplication by zero at those points turned the undefined original into a true statement, manufacturing both roots. A sanity check that the answer is right. Rewrite the original with a common denominator without clearing: \( \dfrac{x(x+2) - 2(x-2)}{(x-2)(x+2)} = \dfrac{x^2 + 4}{x^2 - 4} \). Setting that equal to \( \dfrac{8}{x^2 - 4} \) requires \( x^2 + 4 = 8 \) and \( x^2 - 4 \ne 0 \). The first gives \( x^2 = 4 \) and the second forbids it. The two conditions are incompatible, so no solution exists. This route reaches the same conclusion without ever producing a false candidate. No solution; both candidates are extraneous

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the vertical asymptote of \( \dfrac{1}{x - 4} \).
   Show the full solution

   \( x = 4 \)

2. Find the horizontal asymptote of \( \dfrac{5}{x + 2} \).
   Show the full solution

   Numerator degree 0, denominator degree 1. \( y = 0 \)

3. Find the horizontal asymptote of \( \dfrac{4x}{x - 7} \).
   Show the full solution

   Equal degrees, ratio of leading coefficients. \( y = 4 \)

4. Does \( \dfrac{x - 1}{x - 1} \) have an asymptote or a hole at \( x = 1 \)?
   Show the full solution

   The factor cancels. A hole

5. Find the \( y \)-intercept of \( \dfrac{x + 6}{x - 3} \).
   Show the full solution

   \( \dfrac{6}{-3} \). \( -2 \)

6. Analyze \( f(x) = \dfrac{x - 5}{x^2 - 25} \): holes, asymptotes and intercepts.
   Show the full solution

   Factor: \( \dfrac{x - 5}{(x - 5)(x + 5)} \). Excluded: \( x = 5 \) and \( x = -5 \). \( (x - 5) \) cancels, so \( x = 5 \) is a hole. Simplified: \( \dfrac{1}{x + 5} \). Hole height: \( \dfrac{1}{5 + 5} = \dfrac{1}{10} \), so the hole is at \( \left( 5, \dfrac{1}{10} \right) \). \( (x + 5) \) remains, so \( x = -5 \) is a vertical asymptote. Horizontal asymptote: numerator degree 1, denominator degree 2, so \( y = 0 \). \( x \)-intercept: the simplified numerator is the constant 1, never zero, so there is none. \( y \)-intercept: \( \dfrac{-5}{-25} = \dfrac{1}{5} \). Hole at \( \left( 5, \frac{1}{10} \right) \), vertical asymptote \( x = -5 \), horizontal asymptote \( y = 0 \)

7. Find all asymptotes of \( f(x) = \dfrac{2x^2 + 3}{x^2 - 1} \).
   Show the full solution

   Denominator: \( (x - 1)(x + 1) \), zero at \( x = \pm 1 \). The numerator \( 2x^2 + 3 \) is never zero, so neither factor cancels and both give vertical asymptotes: \( x = 1 \) and \( x = -1 \). Horizontal: degrees equal, so \( y = \dfrac{2}{1} = 2 \). Check at \( x = 100 \): \( \dfrac{20003}{9999} \approx 2.0005 \). Near 2. Correct. Vertical \( x = \pm 1 \), horizontal \( y = 2 \)

8. Find the slant asymptote of \( f(x) = \dfrac{x^2 - 3x + 2}{x + 1} \).
   Show the full solution

   Numerator degree 2, denominator degree 1, so the difference is 1 and a slant asymptote exists. Synthetic division with \( c = -1 \) on \( 1, -3, 2 \): bring down 1; \( -3 - 1 = -4 \); \( 2 + 4 = 6 \). Quotient \( x - 4 \), remainder 6. \( f(x) = x - 4 + \dfrac{6}{x + 1} \). Slant asymptote: \( y = x - 4 \). Check at \( x = 99 \): \( \dfrac{9801 - 297 + 2}{100} = \dfrac{9506}{100} = 95.06 \), and the line gives \( 99 - 4 = 95 \). Within 0.06. Correct. Also a vertical asymptote at \( x = -1 \), since the numerator \( (x-1)(x-2) \) is not zero there. \( y = x - 4 \)

9. Explain why the horizontal asymptote is the ratio of leading coefficients when the degrees match.
   Show the full solution

   Divide the numerator and denominator by the highest power of \( x \) appearing. For \( \dfrac{ax^n + \cdots}{bx^n + \cdots} \), dividing top and bottom by \( x^n \) gives \[ \frac{a + \frac{\cdots}{x} + \cdots}{b + \frac{\cdots}{x} + \cdots} \] Every term except \( a \) and \( b \) now has a positive power of \( x \) in its denominator, so every one approaches zero as \( \left| x \right| \) grows. The quotient approaches \( \dfrac{a}{b} \). The same argument covers the other cases. If the numerator's degree is smaller, dividing by the denominator's leading power leaves the numerator with only vanishing terms, so the limit is \( \dfrac{0}{b} = 0 \). If the numerator's degree is larger, the numerator grows without bound relative to the denominator and no horizontal asymptote exists. Why this is the same idea as end behavior. Lesson 3.1 argued that a polynomial behaves like its leading term far from the origin. A rational function is a ratio of two polynomials, so far from the origin it behaves like the ratio of their leading terms, \( \dfrac{ax^n}{bx^m} \). When \( n = m \) that is the constant \( \dfrac{a}{b} \). A caution about crossing. A graph may cross its horizontal asymptote, unlike a vertical one. The asymptote describes behavior far away, not a barrier. For example \( \dfrac{x}{x^2 + 1} \) has asymptote \( y = 0 \) and crosses it at the origin. Dividing through by the common leading power sends every other term to zero, leaving \( \frac{a}{b} \)

10. Analyze \( f(x) = \dfrac{x^3 - x^2 - 6x}{x^2 - 9} \) completely: holes, all asymptotes, and intercepts.
    Show the full solution

    Factor both parts. Numerator: \( x^3 - x^2 - 6x = x(x^2 - x - 6) = x(x - 3)(x + 2) \). Denominator: \( x^2 - 9 = (x - 3)(x + 3) \). Excluded values. \( x = 3 \) and \( x = -3 \). Classify them. \( (x - 3) \) appears in both and cancels, so \( x = 3 \) is a hole. \( (x + 3) \) remains in the denominator, so \( x = -3 \) is a vertical asymptote. Simplify. \( f(x) = \dfrac{x(x + 2)}{x + 3} = \dfrac{x^2 + 2x}{x + 3} \), with \( x \ne 3 \). Locate the hole. Substitute \( x = 3 \) into the simplified form: \( \dfrac{3(5)}{6} = \dfrac{15}{6} = \dfrac{5}{2} \). Hole at \( \left( 3,\, \dfrac{5}{2} \right) \). Find the slant asymptote. In the simplified form the numerator's degree is 2 and the denominator's is 1, a difference of exactly 1, so a slant asymptote exists. Synthetic division with \( c = -3 \) on \( 1, 2, 0 \): bring down 1; \( 2 - 3 = -1 \); \( 0 + 3 = 3 \). Quotient \( x - 1 \), remainder 3. \( f(x) = x - 1 + \dfrac{3}{x + 3} \). Slant asymptote: \( y = x - 1 \). Check the slant numerically. At \( x = 97 \): \( f(97) = \dfrac{9409 + 194}{100} = \dfrac{9603}{100} = 96.03 \), and the line gives \( 96 \). Within 0.03. Correct. Note that there is no horizontal asymptote, since a function cannot have both. The numerator's degree exceeds the denominator's, so the values grow without bound. Find the intercepts. \( x \)-intercepts from the simplified numerator \( x(x + 2) \): at \( x = 0 \) and \( x = -2 \). Both are legal values, so both are genuine intercepts: \( (0, 0) \) and \( (-2, 0) \). \( y \)-intercept: \( f(0) = \dfrac{0}{3} = 0 \), the same point as one of the \( x \)-intercepts. The graph passes through the origin. Verify the intercepts in the original. \( f(-2) = \dfrac{-8 - 4 + 12}{4 - 9} = \dfrac{0}{-5} = 0 \). Correct. Sign behavior near the vertical asymptote. At \( x = -2.9 \): \( \dfrac{(-2.9)(-0.9)}{0.1} = \dfrac{2.61}{0.1} = 26.1 \), large and positive. At \( x = -3.1 \): \( \dfrac{(-3.1)(-1.1)}{-0.1} = \dfrac{3.41}{-0.1} = -34.1 \), large and negative. So the graph rises to \( +\infty \) on the right of \( x = -3 \) and falls to \( -\infty \) on the left, which is the behavior of an odd-order asymptote. The full description. Hole at \( \left( 3, \frac{5}{2} \right) \), vertical asymptote at \( x = -3 \), slant asymptote \( y = x - 1 \), intercepts at \( (0,0) \) and \( (-2, 0) \). What the analysis order buys. Factoring first and canceling second is what separated the hole from the asymptote. Simplifying first would have produced the right slant asymptote but lost the hole entirely, which is the error lesson 4.1 warned about and this problem is built to expose. Hole \( \left( 3, \frac{5}{2} \right) \); vertical asymptote \( x = -3 \); slant asymptote \( y = x - 1 \); intercepts \( (0,0) \) and \( (-2,0) \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A worker finishes a job in 5 hours. What fraction does he do per hour?
   Show the full solution

   \( \frac{1}{5} \)

2. Two workers take 4 and 4 hours alone. How long together?
   Show the full solution

   Rate \( \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2} \). 2 hours

3. How much salt is in 20 L of a 15 percent solution?
   Show the full solution

   \( 0.15 \times 20 \). 3 L

4. Total cost is \( 300 + 5x \). Write the average cost.
   Show the full solution

   \( \frac{300}{x} + 5 \)

5. What is that function's horizontal asymptote?
   Show the full solution

   \( y = 5 \)

6. Pipe A fills a pool in 9 hours and pipe B in 18 hours. How long together?
   Show the full solution

   Combined rate: \( \dfrac{1}{9} + \dfrac{1}{18} = \dfrac{2}{18} + \dfrac{1}{18} = \dfrac{3}{18} = \dfrac{1}{6} \). Time: 6 hours. Check: in 6 hours A does \( \dfrac{6}{9} = \dfrac{2}{3} \) and B does \( \dfrac{6}{18} = \dfrac{1}{3} \). Total 1. Correct. Plausibility: 6 hours is less than 9, the faster pipe alone. Correct. 6 hours

7. How much pure alcohol must be added to 40 L of a 20 percent solution to make it 50 percent?
   Show the full solution

   Current alcohol: \( 0.20 \times 40 = 8 \) L. Adding \( x \) L of pure alcohol adds \( x \) L of alcohol and \( x \) L to the total. New concentration: \( \dfrac{8 + x}{40 + x} = 0.50 \). Multiply: \( 8 + x = 0.5(40 + x) = 20 + 0.5x \). \( 0.5x = 12 \), so \( x = 24 \) L. Check: alcohol \( 8 + 24 = 32 \), total \( 40 + 24 = 64 \), and \( \dfrac{32}{64} = 0.50 \). Correct. Note the difference from the worked example: adding pure solute changes both the numerator and the denominator, while adding pure water changed only the denominator. 24 L

8. A car travels 120 km at speed \( v \) and returns at \( v + 20 \). Write the total time as a function of \( v \) and find it when \( v = 60 \).
   Show the full solution

   Time is distance over speed. Out: \( \dfrac{120}{v} \). Back: \( \dfrac{120}{v + 20} \). Total: \( T(v) = \dfrac{120}{v} + \dfrac{120}{v + 20} \). Combined over a common denominator: \( T(v) = \dfrac{120(v + 20) + 120v}{v(v + 20)} = \dfrac{240v + 2400}{v(v + 20)} \). At \( v = 60 \): \( \dfrac{120}{60} + \dfrac{120}{80} = 2 + 1.5 = 3.5 \) hours. Check with the combined form: \( \dfrac{14400 + 2400}{60(80)} = \dfrac{16800}{4800} = 3.5 \). Agrees. Domain: \( v \gt 0 \), since a speed must be positive. \( T(v) = \frac{240v + 2400}{v(v+20)} \); 3.5 hours at \( v = 60 \)

9. Explain why average cost always exceeds the variable cost per unit.
   Show the full solution

   Write the total cost as fixed plus variable: \( C(x) = F + vx \), with \( F \gt 0 \) the fixed cost and \( v \) the variable cost per unit. Average cost is \( \overline{C}(x) = \dfrac{F + vx}{x} = \dfrac{F}{x} + v \). Since \( F \gt 0 \) and \( x \gt 0 \), the term \( \dfrac{F}{x} \) is strictly positive. So \( \overline{C}(x) \gt v \) for every positive \( x \). What that means economically. Every unit produced carries its own variable cost plus a share of the fixed cost, and the share is never zero. So the average can approach the variable cost but never reach it. The graph. The horizontal asymptote at \( y = v \) is approached from above and never crossed, which is unusual: many rational functions do cross their horizontal asymptotes, and this one cannot for a structural reason rather than an accidental one. The limiting case. If the fixed cost were zero, average cost would equal variable cost exactly at every volume, and the function would be the constant \( v \) rather than a rational function at all. The curvature in the graph is entirely caused by the fixed cost. Why firms care. The gap \( \dfrac{F}{x} \) is the per-unit burden of the fixed cost, and it is what makes low-volume production expensive. Halving it requires doubling output, which is why the savings from scale diminish so quickly. Average cost is \( v + \frac{F}{x} \), and the second term is strictly positive for any positive output

10. Working together, two crews resurface a road in 6 days. Alone, the first crew would take 5 days longer than the second. Find how long each takes alone.
    Show the full solution

    Define the variable. Let \( x \) be the number of days the second crew takes alone. Then the first takes \( x + 5 \) days. Write the rates. Second crew: \( \dfrac{1}{x} \) of the job per day. First crew: \( \dfrac{1}{x + 5} \) per day. Together they complete the job in 6 days, so their combined rate is \( \dfrac{1}{6} \) per day. Set up the equation. \( \dfrac{1}{x} + \dfrac{1}{x + 5} = \dfrac{1}{6} \). Note the exclusions. \( x \ne 0 \) and \( x \ne -5 \). Clear the denominators by multiplying every term by \( 6x(x + 5) \): \( 6(x + 5) + 6x = x(x + 5) \). Expand and solve. \( 6x + 30 + 6x = x^2 + 5x \). \( 12x + 30 = x^2 + 5x \). \( 0 = x^2 - 7x - 30 \). Factor: two numbers multiplying to \( -30 \) and adding to \( -7 \) are \( -10 \) and 3: \( 0 = (x - 10)(x + 3) \). Candidates: \( x = 10 \) and \( x = -3 \). Apply the checks. Neither is an excluded value, so neither is extraneous in the algebraic sense. But \( x = -3 \) is rejected by the situation: a crew cannot take a negative number of days. The domain the problem imposes is \( x \gt 0 \). Verify \( x = 10 \) in the original. Second crew: 10 days, rate \( \dfrac{1}{10} \). First crew: 15 days, rate \( \dfrac{1}{15} \). Combined: \( \dfrac{1}{10} + \dfrac{1}{15} = \dfrac{3}{30} + \dfrac{2}{30} = \dfrac{5}{30} = \dfrac{1}{6} \). Correct, giving 6 days together. Check the work done. In 6 days the second crew does \( \dfrac{6}{10} = 0.6 \) of the job and the first does \( \dfrac{6}{15} = 0.4 \). Together exactly 1. Correct. Plausibility check. The combined time of 6 days is less than 10, the faster crew's solo time, as it must be. And the slower crew at 15 days is indeed 5 days longer than the faster. Both conditions hold. The distinction worth noting. The root \( x = -3 \) was not extraneous in the technical sense of lesson 4.5: it genuinely satisfies the cleared equation, and substituting it into the original gives \( -\dfrac{1}{3} + \dfrac{1}{2} = \dfrac{1}{6} \), which is true. It was rejected by the context, not by the algebra. Distinguishing "extraneous" from "rejected by the situation" is worth keeping straight, because the two require different justifications in a written answer. The second crew takes 10 days alone and the first takes 15

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Write \( \sqrt{x} \) with a rational exponent.
   Show the full solution

   \( x^{1/2} \)

2. Write \( \sqrt[3]{x^2} \) with a rational exponent.
   Show the full solution

   \( x^{2/3} \)

3. Evaluate \( 9^{1/2} \).
   Show the full solution

   3

4. Evaluate \( 8^{1/3} \).
   Show the full solution

   2

5. Evaluate \( 4^{-1/2} \).
   Show the full solution

   \( \dfrac{1}{4^{1/2}} = \dfrac{1}{2} \). \( \frac{1}{2} \)

6. Evaluate \( 32^{3/5} \).
   Show the full solution

   Root first: \( \sqrt[5]{32} = 2 \), since \( 2^5 = 32 \). Then cube: \( 2^3 = 8 \). Check the other order: \( \sqrt[5]{32^3} = \sqrt[5]{32768} = 8 \), since \( 8^5 = 32768 \). Agrees, but the numbers were much larger. 8

7. Evaluate \( \left( \dfrac{1}{16} \right)^{-3/4} \).
   Show the full solution

   The negative exponent inverts the base: \( \left( \dfrac{1}{16} \right)^{-3/4} = 16^{3/4} \). Root first: \( \sqrt[4]{16} = 2 \). Then cube: \( 8 \). Check: \( \left( \dfrac{1}{16} \right)^{3/4} = \dfrac{1}{8} \), and its reciprocal is 8. Agrees. 8

8. Simplify \( \left( x^{2/3} y^{-1/2} \right)^6 \).
   Show the full solution

   Raise each factor to the sixth, multiplying exponents: \( x^{(2/3)(6)} y^{(-1/2)(6)} = x^4 y^{-3} \). Write with a positive exponent: \( \dfrac{x^4}{y^3} \). Check at \( x = 8 \), \( y = 4 \): \( x^{2/3} = 4 \), \( y^{-1/2} = \dfrac{1}{2} \), so the inside is 2 and \( 2^6 = 64 \). The answer gives \( \dfrac{4096}{64} = 64 \). Agrees. \( \frac{x^4}{y^3} \)

9. Explain why \( x^{1/n} \) has to mean the \( n \)th root rather than anything else.
   Show the full solution

   The exponent rules were established for integers, and extending them to fractions is only worth doing if the rules keep holding. That requirement determines the meaning completely. The argument. If \( (x^a)^b = x^{ab} \) is to survive, then \( \left( x^{1/n} \right)^n = x^{(1/n)(n)} = x^1 = x \). So \( x^{1/n} \) is a quantity whose \( n \)th power is \( x \), which is the definition of an \( n \)th root. Why no other choice works. Any other definition would break the power rule, and then every algebraic manipulation involving exponents would need a separate case for fractions. The whole value of the notation is that one set of rules covers integers and fractions alike. The same reasoning fixes other extensions. \( x^0 = 1 \) because \( x^a x^0 = x^{a+0} = x^a \) forces \( x^0 \) to be the multiplicative identity. And \( x^{-n} = \dfrac{1}{x^n} \) because \( x^n x^{-n} = x^0 = 1 \). In each case the definition is the unique one that preserves the rules. What this says about mathematics generally. Definitions of this kind are not arbitrary conventions. They are the only options consistent with what is already established, which is why they feel inevitable once the reason is seen. It is the only definition under which \( (x^a)^b = x^{ab} \) continues to hold

10. Simplify \( \dfrac{\sqrt[3]{x^4} \cdot \sqrt{x}}{\sqrt[6]{x}} \), stating any assumption.
    Show the full solution

    Convert everything to exponent form. \( \sqrt[3]{x^4} = x^{4/3} \). \( \sqrt{x} = x^{1/2} \). \( \sqrt[6]{x} = x^{1/6} \). Combine the exponents. \( x^{4/3 + 1/2 - 1/6} \). Over a common denominator of 6: \( \dfrac{8}{6} + \dfrac{3}{6} - \dfrac{1}{6} = \dfrac{10}{6} = \dfrac{5}{3} \). So the expression is \( x^{5/3} \). Convert back to radical form if wanted. \( x^{5/3} = \sqrt[3]{x^5} = x\sqrt[3]{x^2} \), since \( x^5 = x^3 \cdot x^2 \) and the \( x^3 \) comes out of the cube root as \( x \). State the assumption. The expression contains \( \sqrt{x} \), an even root, so \( x \ge 0 \) is required for a real result. The other two radicals would accept negatives, but the square root does not, so the domain is \( x \ge 0 \). Verify at \( x = 64 \). \( \sqrt[3]{64^4} \): since \( \sqrt[3]{64} = 4 \), this is \( 4^4 = 256 \). \( \sqrt{64} = 8 \). \( \sqrt[6]{64} = 2 \), since \( 2^6 = 64 \). Original: \( \dfrac{256 \times 8}{2} = \dfrac{2048}{2} = 1024 \). Answer: \( 64^{5/3} = \left( \sqrt[3]{64} \right)^5 = 4^5 = 1024 \). Agrees. Verify at \( x = 1 \). Everything is 1, and \( 1^{5/3} = 1 \). Agrees trivially, which is a weak check, but the \( x = 64 \) case is strong. Why exponent form was essential. In radical form, with three different indices, there is no way to combine the terms directly. Converting turns the problem into adding three fractions, which is routine. This is the single most useful reason to learn rational exponents. \( x^{5/3} = x\sqrt[3]{x^2} \), assuming \( x \ge 0 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( \sqrt{50} \).
   Show the full solution

   \( \sqrt{25}\sqrt{2} \). \( 5\sqrt{2} \)

2. Simplify \( \sqrt{48} \).
   Show the full solution

   \( \sqrt{16}\sqrt{3} \). \( 4\sqrt{3} \)

3. Simplify \( \sqrt[3]{16} \).
   Show the full solution

   \( 16 = 8 \times 2 \). \( 2\sqrt[3]{2} \)

4. Simplify \( \sqrt{x^6} \) for \( x \ge 0 \).
   Show the full solution

   \( x^3 \)

5. Simplify \( \sqrt{x^2} \) with no assumption on \( x \).
   Show the full solution

   \( \left| x \right| \)

6. Simplify \( \sqrt{75x^4y^3} \), assuming \( y \ge 0 \).
   Show the full solution

   Factor: \( 75 = 25 \times 3 \); \( x^4 \) is a perfect square; \( y^3 = y^2 \cdot y \). Extract: \( \sqrt{25} = 5 \), \( \sqrt{x^4} = x^2 \) with no absolute value needed since \( x^2 \ge 0 \) always, and \( \sqrt{y^2} = y \) given \( y \ge 0 \). \( 5x^2y\sqrt{3y} \). Check at \( x = 1 \), \( y = 3 \): radicand \( 75(1)(27) = 2025 \), and \( \sqrt{2025} = 45 \). Answer: \( 5(1)(3)\sqrt{9} = 15 \times 3 = 45 \). Correct. \( 5x^2y\sqrt{3y} \)

7. Simplify \( \sqrt[3]{-40x^5} \).
   Show the full solution

   An odd root accepts a negative radicand, so no restriction is needed. Factor: \( -40 = -8 \times 5 \), and \( -8 \) is a perfect cube. \( x^5 = x^3 \cdot x^2 \). Extract: \( \sqrt[3]{-8} = -2 \) and \( \sqrt[3]{x^3} = x \), with no absolute value. \( -2x\sqrt[3]{5x^2} \). Check at \( x = 1 \): radicand \( -40 \), and \( \sqrt[3]{-40} \approx -3.4200 \). Answer: \( -2\sqrt[3]{5} \approx -2(1.70998) \approx -3.4200 \). Correct. \( -2x\sqrt[3]{5x^2} \)

8. Simplify \( \sqrt{\dfrac{9x^4}{25y^2}} \), assuming \( y \gt 0 \).
   Show the full solution

   Apply the quotient property: \( \dfrac{\sqrt{9x^4}}{\sqrt{25y^2}} \). Numerator: \( \sqrt{9} = 3 \) and \( \sqrt{x^4} = x^2 \), nonnegative automatically. Denominator: \( \sqrt{25} = 5 \) and \( \sqrt{y^2} = y \), given \( y \gt 0 \). \( \dfrac{3x^2}{5y} \). Check at \( x = 2 \), \( y = 1 \): radicand \( \dfrac{9(16)}{25} = \dfrac{144}{25} \), and its root is \( \dfrac{12}{5} \). Answer: \( \dfrac{3(4)}{5} = \dfrac{12}{5} \). Correct. \( \frac{3x^2}{5y} \)

9. Explain why \( \sqrt{x^2} = \left| x \right| \) but \( \sqrt[3]{x^3} = x \).
   Show the full solution

   The difference is whether the power destroys sign information. The square case. Squaring maps both 5 and \( -5 \) to 25. The radical sign denotes the principal root, which is by convention the nonnegative one, so \( \sqrt{25} = 5 \). Starting from \( x = -5 \), squaring gives 25 and rooting gives 5, which is \( \left| x \right| \) rather than \( x \). The sign was lost in the squaring and the root cannot recover it. The cube case. Cubing preserves sign: \( (-5)^3 = -125 \) and \( 5^3 = 125 \). The cube root is therefore unambiguous, with \( \sqrt[3]{-125} = -5 \). Starting from \( x = -5 \), cubing gives \( -125 \) and rooting returns \( -5 \), which is \( x \) exactly. No information was lost, so no absolute value is needed. The general statement. For even \( n \), \( \sqrt[n]{x^n} = \left| x \right| \). For odd \( n \), \( \sqrt[n]{x^n} = x \). Where the same distinction recurs. Even roots have restricted domains and odd roots do not. \( x^2 \) needs a restricted domain to have an inverse and \( x^3 \) does not. The parity of the exponent runs through every part of this unit, and it is always the same underlying fact. Even powers destroy sign, so the principal root returns the magnitude; odd powers preserve it, so the root recovers the original

10. Simplify \( \sqrt{4x^2 + 12x + 9} \) with no assumption on \( x \), and evaluate it at \( x = 1 \) and \( x = -4 \).
    Show the full solution

    Factor the radicand. Check whether it is a perfect square trinomial. The first term is \( (2x)^2 \) and the last is \( 3^2 \), so the middle term would need to be \( 2(2x)(3) = 12x \). It is. \( 4x^2 + 12x + 9 = (2x + 3)^2 \). Take the root. The index is even and nothing is assumed about \( x \), so \( 2x + 3 \) could be negative: \( \sqrt{(2x + 3)^2} = \left| 2x + 3 \right| \). Evaluate at \( x = 1 \). Directly: the radicand is \( 4 + 12 + 9 = 25 \), so the value is 5. From the answer: \( \left| 2 + 3 \right| = 5 \). Agrees. Note that \( 2x + 3 = 5 \) here, positive, so the absolute value changed nothing. Evaluate at \( x = -4 \). Directly: the radicand is \( 4(16) + 12(-4) + 9 = 64 - 48 + 9 = 25 \), so the value is 5. From the answer: \( \left| -8 + 3 \right| = \left| -5 \right| = 5 \). Agrees. This is the case that proves the absolute value necessary. Without it the answer would have been \( 2x + 3 = -5 \), which is wrong: a principal square root is never negative. Find where the sign changes. \( 2x + 3 \) is negative when \( x \lt -\dfrac{3}{2} \). So the expression equals \( 2x + 3 \) for \( x \ge -\dfrac{3}{2} \) and \( -(2x + 3) \) for \( x \lt -\dfrac{3}{2} \). Writing it piecewise makes the behavior explicit: \[ \sqrt{4x^2 + 12x + 9} = \begin{cases} 2x + 3 & x \ge -\frac{3}{2} \\ -2x - 3 & x \lt -\frac{3}{2} \end{cases} \] The shape this describes. The graph is a V with its corner at \( \left( -\dfrac{3}{2},\, 0 \right) \), which is the absolute value parent stretched by 2 and shifted. That corner is exactly where the radicand's perfect square touches zero, and it is why an expression that looks like a smooth parabola under a radical comes out with a sharp point. \( \left| 2x + 3 \right| \); the value is 5 at both \( x = 1 \) and \( x = -4 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( 2\sqrt{5} + 7\sqrt{5} \).
   Show the full solution

   \( 9\sqrt{5} \)

2. Can \( \sqrt{2} + \sqrt{5} \) be combined?
   Show the full solution

   Different radicands. No

3. Simplify \( \sqrt{3} \cdot \sqrt{12} \).
   Show the full solution

   \( \sqrt{36} \). 6

4. Simplify \( \left( \sqrt{11} \right)^2 \).
   Show the full solution

   11

5. Simplify \( \sqrt{8} + \sqrt{18} \).
   Show the full solution

   \( 2\sqrt{2} + 3\sqrt{2} \). \( 5\sqrt{2} \)

6. Simplify \( 4\sqrt{20} - 2\sqrt{45} + \sqrt{5} \).
   Show the full solution

   Simplify each: \( \sqrt{20} = 2\sqrt{5} \) and \( \sqrt{45} = 3\sqrt{5} \). \( 4(2\sqrt{5}) - 2(3\sqrt{5}) + \sqrt{5} = 8\sqrt{5} - 6\sqrt{5} + \sqrt{5} = 3\sqrt{5} \). Check: \( 4(4.4721) - 2(6.7082) + 2.2361 \approx 17.889 - 13.416 + 2.236 = 6.708 \), and \( 3\sqrt{5} \approx 6.708 \). Correct. \( 3\sqrt{5} \)

7. Expand \( (3 - \sqrt{5})^2 \).
   Show the full solution

   \( (a - b)^2 = a^2 - 2ab + b^2 \) with \( a = 3 \), \( b = \sqrt{5} \). \( 9 - 2(3)\sqrt{5} + 5 = 14 - 6\sqrt{5} \). Check: \( (3 - 2.2361)^2 = (0.7639)^2 \approx 0.5835 \), and \( 14 - 6(2.2361) \approx 14 - 13.4164 = 0.5836 \). Correct. \( 14 - 6\sqrt{5} \)

8. Expand \( (2\sqrt{3} + \sqrt{2})(\sqrt{3} - 3\sqrt{2}) \).
   Show the full solution

   Four products: \( 2\sqrt{3} \cdot \sqrt{3} = 2(3) = 6 \). \( 2\sqrt{3} \cdot (-3\sqrt{2}) = -6\sqrt{6} \). \( \sqrt{2} \cdot \sqrt{3} = \sqrt{6} \). \( \sqrt{2} \cdot (-3\sqrt{2}) = -3(2) = -6 \). Collect: rational \( 6 - 6 = 0 \); radical \( -6\sqrt{6} + \sqrt{6} = -5\sqrt{6} \). Answer: \( -5\sqrt{6} \). Check: \( (3.4641 + 1.4142)(1.7321 - 4.2426) = (4.8783)(-2.5105) \approx -12.247 \), and \( -5\sqrt{6} \approx -5(2.4495) = -12.247 \). Correct. \( -5\sqrt{6} \)

9. Explain why simplifying radicals before adding is necessary rather than optional.
   Show the full solution

   Two radicals can be like without looking like it. The test for like radicals is applied to the simplified forms, not to whatever was written down. The worked case. \( \sqrt{12} \) and \( \sqrt{27} \) have different radicands and appear uncombinable. Simplified they are \( 2\sqrt{3} \) and \( 3\sqrt{3} \), plainly like, and they combine to \( 5\sqrt{3} \). A student who skipped the simplification would report \( \sqrt{12} + \sqrt{27} \) as already in simplest form, which is wrong on two counts: the terms are not simplified and they do combine. It also works the other way. Simplification can reveal that terms which look combinable are not. \( \sqrt{8} \) and \( \sqrt{12} \) both simplify, to \( 2\sqrt{2} \) and \( 2\sqrt{3} \), and those are genuinely unlike. Without simplifying, a careless reader might combine the coefficients. The reliable procedure. Simplify every radical completely first, then compare index and radicand, then combine only exact matches. Skipping the first step makes the second step unreliable. A parallel with fractions. \( \dfrac{2}{6} + \dfrac{1}{3} \) looks like unlike fractions until the first is reduced to \( \dfrac{1}{3} \). The situation is identical: simplification is what makes the comparison meaningful. Radicals that look unlike often become like after simplification, so the comparison is only valid on simplified forms

10. Simplify \( \sqrt{50x^3} - x\sqrt{18x} + \sqrt{8x^3} \), assuming \( x \ge 0 \), and verify at \( x = 2 \).
    Show the full solution

    Simplify each term. First: \( \sqrt{50x^3} = \sqrt{25 \cdot 2 \cdot x^2 \cdot x} = 5x\sqrt{2x} \). Second: \( x\sqrt{18x} = x\sqrt{9 \cdot 2 \cdot x} = 3x\sqrt{2x} \). Third: \( \sqrt{8x^3} = \sqrt{4 \cdot 2 \cdot x^2 \cdot x} = 2x\sqrt{2x} \). All three are like radicals, each a multiple of \( \sqrt{2x} \), although none looked that way at the start. Combine the coefficients. \( 5x - 3x + 2x = 4x \). Answer: \( 4x\sqrt{2x} \). Verify at \( x = 2 \). First term: \( \sqrt{50(8)} = \sqrt{400} = 20 \). Second term: \( 2\sqrt{36} = 2(6) = 12 \). Third term: \( \sqrt{8(8)} = \sqrt{64} = 8 \). Original: \( 20 - 12 + 8 = 16 \). Answer: \( 4(2)\sqrt{4} = 8 \times 2 = 16 \). Agrees. Verify at \( x = 1 \). Original: \( \sqrt{50} - \sqrt{18} + \sqrt{8} \approx 7.0711 - 4.2426 + 2.8284 = 5.6569 \). Answer: \( 4\sqrt{2} \approx 5.6569 \). Agrees. Why the assumption was needed. Each simplification used \( \sqrt{x^2} = x \), which requires \( x \ge 0 \). Without it the answer would be \( 4\left| x \right|\sqrt{2x} \), and in fact the whole expression requires \( x \ge 0 \) anyway for the radicands to be nonnegative, so the assumption costs nothing here. The pattern worth extracting. Every term reduced to a multiple of the same radical \( \sqrt{2x} \). That is common when the radicands share the same non-square part, here \( 2x \), and differ only by perfect square factors. Spotting that in advance predicts that the terms will combine. \( 4x\sqrt{2x} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Rationalize \( \dfrac{1}{\sqrt{2}} \).
   Show the full solution

   \( \frac{\sqrt{2}}{2} \)

2. Rationalize \( \dfrac{4}{\sqrt{5}} \).
   Show the full solution

   \( \frac{4\sqrt{5}}{5} \)

3. Give the conjugate of \( 2 + \sqrt{7} \).
   Show the full solution

   \( 2 - \sqrt{7} \)

4. Compute \( (4 - \sqrt{3})(4 + \sqrt{3}) \).
   Show the full solution

   \( 16 - 3 \). 13

5. Rationalize \( \dfrac{3}{\sqrt{12}} \).
   Show the full solution

   Simplify first: \( \sqrt{12} = 2\sqrt{3} \), so the expression is \( \dfrac{3}{2\sqrt{3}} \). Multiply by \( \dfrac{\sqrt{3}}{\sqrt{3}} \): \( \dfrac{3\sqrt{3}}{6} = \dfrac{\sqrt{3}}{2} \). \( \frac{\sqrt{3}}{2} \)

6. Rationalize \( \dfrac{2}{\sqrt{5} + 1} \).
   Show the full solution

   Multiply by the conjugate \( \sqrt{5} - 1 \): Denominator: \( 5 - 1 = 4 \). Numerator: \( 2(\sqrt{5} - 1) = 2\sqrt{5} - 2 \). \( \dfrac{2\sqrt{5} - 2}{4} = \dfrac{2(\sqrt{5} - 1)}{4} = \dfrac{\sqrt{5} - 1}{2} \). Check: \( \dfrac{2}{3.2361} \approx 0.6180 \), and \( \dfrac{2.2361 - 1}{2} = 0.6180 \). Correct. \( \frac{\sqrt{5} - 1}{2} \)

7. Rationalize \( \dfrac{\sqrt{3}}{\sqrt{6} - \sqrt{3}} \).
   Show the full solution

   Conjugate: \( \sqrt{6} + \sqrt{3} \). Denominator: \( 6 - 3 = 3 \). Numerator: \( \sqrt{3}(\sqrt{6} + \sqrt{3}) = \sqrt{18} + 3 = 3\sqrt{2} + 3 \). \( \dfrac{3\sqrt{2} + 3}{3} = \dfrac{3(\sqrt{2} + 1)}{3} = \sqrt{2} + 1 \). Check: \( \dfrac{1.7321}{2.4495 - 1.7321} = \dfrac{1.7321}{0.7174} \approx 2.4143 \), and \( 1.4142 + 1 = 2.4142 \). Correct. \( \sqrt{2} + 1 \)

8. Rationalize \( \dfrac{1}{\sqrt[3]{4}} \).
   Show the full solution

   The square-root trick does not apply. Multiplying by \( \sqrt[3]{4} \) gives \( \sqrt[3]{16} \) underneath, which is still a radical. Find what completes the cube. \( 4 = 2^2 \), so \( \sqrt[3]{4} = \sqrt[3]{2^2} = 2^{2/3} \). Multiplying by \( 2^{1/3} \) gives \( 2^{3/3} = 2 \), a whole number. And \( 2^{1/3} = \sqrt[3]{2} \). Multiply top and bottom by \( \sqrt[3]{2} \): \( \dfrac{\sqrt[3]{2}}{\sqrt[3]{4} \cdot \sqrt[3]{2}} = \dfrac{\sqrt[3]{2}}{\sqrt[3]{8}} = \dfrac{\sqrt[3]{2}}{2} \). Check: \( \dfrac{1}{1.5874} \approx 0.6300 \), and \( \dfrac{1.2599}{2} = 0.6300 \). Correct. The general rule: for \( \sqrt[n]{a^m} \) in a denominator, multiply by \( \sqrt[n]{a^{n-m}} \) so the exponents sum to \( n \). \( \frac{\sqrt[3]{2}}{2} \)

9. Explain why rationalizing does not change the value of an expression.
   Show the full solution

   Every step multiplies the expression by a fraction whose numerator and denominator are identical, which is multiplication by 1. Multiplying by 1 leaves a value unchanged. Written out. \( \dfrac{A}{B} = \dfrac{A}{B} \cdot 1 = \dfrac{A}{B} \cdot \dfrac{k}{k} = \dfrac{Ak}{Bk} \), valid for any \( k \ne 0 \). The multiplier is chosen so that \( Bk \) is rational, but any nonzero \( k \) would preserve the value; the choice is about the form, not the number. The one condition. The multiplier must be nonzero. A conjugate \( a - b \) is zero only when \( a = b \), which would mean the original denominator was also zero and the expression was undefined to begin with. So the condition is automatically satisfied whenever the expression makes sense. Confirmed numerically throughout. Every worked example in this lesson checked the decimal value before and after, and they matched each time. That is not a coincidence but a consequence of multiplying by 1. What changes and what does not. The appearance changes, the value does not. \( \dfrac{1}{\sqrt{2}} \) and \( \dfrac{\sqrt{2}}{2} \) are the same number, approximately 0.7071, written two ways. Neither is more correct; the second is the conventional form. It multiplies by \( \frac{k}{k} \), which equals 1, so only the form changes

10. Rationalize \( \dfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} - \sqrt{y}} \) for positive \( x \) and \( y \) with \( x \ne y \), and verify at \( x = 9 \), \( y = 4 \).
    Show the full solution

    Multiply by the conjugate \( \sqrt{x} + \sqrt{y} \). \( \dfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} - \sqrt{y}} \cdot \dfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} + \sqrt{y}} \). Denominator. \( (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) = x - y \), a difference of squares with both radicals eliminated. Numerator. \( (\sqrt{x} + \sqrt{y})^2 = x + 2\sqrt{xy} + y \). The answer. \( \dfrac{x + y + 2\sqrt{xy}}{x - y} \). Why \( x \ne y \) was required. If \( x = y \) the denominator of the original is zero, so the expression is undefined. The condition was stated in the problem and the answer's denominator \( x - y \) shows why. Verify at \( x = 9 \), \( y = 4 \). Original: \( \dfrac{3 + 2}{3 - 2} = \dfrac{5}{1} = 5 \). Answer: \( \dfrac{9 + 4 + 2\sqrt{36}}{9 - 4} = \dfrac{13 + 12}{5} = \dfrac{25}{5} = 5 \). Agrees. Verify at \( x = 16 \), \( y = 1 \). Original: \( \dfrac{4 + 1}{4 - 1} = \dfrac{5}{3} \approx 1.6667 \). Answer: \( \dfrac{16 + 1 + 2\sqrt{16}}{16 - 1} = \dfrac{17 + 8}{15} = \dfrac{25}{15} = \dfrac{5}{3} \). Agrees. Verify at \( x = 2 \), \( y = 1 \), where the roots are irrational. Original: \( \dfrac{1.41421 + 1}{1.41421 - 1} = \dfrac{2.41421}{0.41421} \approx 5.8284 \). Answer: \( \dfrac{2 + 1 + 2\sqrt{2}}{2 - 1} = 3 + 2(1.41421) = 5.8284 \). Agrees. A structural observation. Note that \( \sqrt{xy} \) survives in the numerator. Rationalizing removes radicals from the denominator only; it makes no promise about the numerator, and here a new radical appeared there. That is normal and is not a failure of the method. The pattern. The answer has the form \( \dfrac{(\sqrt{x} + \sqrt{y})^2}{x - y} \), which is worth recognizing: whenever a fraction's numerator and denominator are conjugates, the rationalized form is the numerator squared over the difference of the squares. \( \frac{x + y + 2\sqrt{xy}}{x - y} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Check every candidate.

1. Solve \( \sqrt{x} = 6 \).
   Show the full solution

   \( x = 36 \)

2. Solve \( \sqrt{x - 3} = 4 \).
   Show the full solution

   \( x - 3 = 16 \). \( x = 19 \)

3. Solve \( \sqrt{x} + 2 = 7 \).
   Show the full solution

   Isolate first: \( \sqrt{x} = 5 \). \( x = 25 \)

4. Solve \( \sqrt[3]{x} = -2 \).
   Show the full solution

   Cube both sides. An odd root accepts a negative value. \( x = -8 \)

5. Solve \( \sqrt{x} = -3 \).
   Show the full solution

   A principal square root is never negative. No solution

6. Solve \( \sqrt{3x + 1} = x - 1 \).
   Show the full solution

   Square: \( 3x + 1 = x^2 - 2x + 1 \). \( 0 = x^2 - 5x = x(x - 5) \). Candidates: \( x = 0 \) and \( x = 5 \). Check \( x = 0 \): left \( \sqrt{1} = 1 \); right \( -1 \). Not equal. Extraneous. Check \( x = 5 \): left \( \sqrt{16} = 4 \); right \( 4 \). Equal. Genuine. \( x = 5 \) only

7. Solve \( \sqrt{x + 5} = \sqrt{2x - 1} \).
   Show the full solution

   Square both sides: \( x + 5 = 2x - 1 \). \( 6 = x \). Check: left \( \sqrt{11} \approx 3.3166 \); right \( \sqrt{11} \approx 3.3166 \). Equal. Genuine. Both radicands are positive at \( x = 6 \), so the domain is satisfied. \( x = 6 \)

8. Solve \( \sqrt{x + 4} + 2 = x \).
   Show the full solution

   Isolate: \( \sqrt{x + 4} = x - 2 \). Square: \( x + 4 = x^2 - 4x + 4 \). \( 0 = x^2 - 5x = x(x - 5) \). Candidates: \( x = 0 \) and \( x = 5 \). Check \( x = 0 \): \( \sqrt{4} + 2 = 2 + 2 = 4 \), but \( x = 0 \). Not equal. Extraneous. Check \( x = 5 \): \( \sqrt{9} + 2 = 3 + 2 = 5 \), and \( x = 5 \). Equal. Genuine. \( x = 5 \) only

9. Explain why the check must compare both sides rather than only test the domain.
   Show the full solution

   There are two distinct ways a candidate can fail, and a domain test catches only one. Failure one: the radicand goes negative. A candidate making \( x - 3 \) negative in \( \sqrt{x - 3} \) is outside the equation's domain, so it cannot be a solution. A domain test catches this. Failure two: the signs disagree. A principal square root is nonnegative, so an equation \( \sqrt{A} = B \) requires \( B \ge 0 \). A candidate making \( B \) negative fails even though every radicand is perfectly legal. In the worked example, \( x = -2 \) gave a radicand of 1, entirely within the domain, but made the right side \( -1 \). A domain test would have passed it. Why squaring hides the second failure. Squaring maps \( 1 \) and \( -1 \) to the same value, so the squared equation cannot distinguish them. The information about sign is destroyed by the operation and can only be recovered by returning to the original. The reliable procedure. Substitute each candidate into the original equation and evaluate both sides independently. If they are equal, the candidate is genuine. That single test catches both failure modes without needing to classify them. A domain test misses candidates whose radicands are legal but whose two sides have opposite signs

10. Solve \( \sqrt{2x + 3} - \sqrt{x - 2} = 2 \), checking every candidate.
    Show the full solution

    Note the domain first. Both radicands must be nonnegative: \( 2x + 3 \ge 0 \) gives \( x \ge -\dfrac{3}{2} \), and \( x - 2 \ge 0 \) gives \( x \ge 2 \). The binding condition is \( x \ge 2 \). Isolate one radical. \( \sqrt{2x + 3} = 2 + \sqrt{x - 2} \). Square both sides. Left: \( 2x + 3 \). Right: \( (2 + \sqrt{x-2})^2 = 4 + 4\sqrt{x - 2} + (x - 2) = x + 2 + 4\sqrt{x - 2} \). So \( 2x + 3 = x + 2 + 4\sqrt{x - 2} \). Isolate the remaining radical. \( x + 1 = 4\sqrt{x - 2} \). Square again. \( (x + 1)^2 = 16(x - 2) \). \( x^2 + 2x + 1 = 16x - 32 \). \( x^2 - 14x + 33 = 0 \). Solve. Two numbers multiplying to 33 and adding to \( -14 \) are \( -3 \) and \( -11 \): \( (x - 3)(x - 11) = 0 \). Candidates: \( x = 3 \) and \( x = 11 \). Both are in the domain \( x \ge 2 \), so the domain test passes both and the substitution check is essential. Check \( x = 3 \). \( \sqrt{6 + 3} - \sqrt{3 - 2} = \sqrt{9} - \sqrt{1} = 3 - 1 = 2 \). Correct. Genuine. Check \( x = 11 \). \( \sqrt{22 + 3} - \sqrt{11 - 2} = \sqrt{25} - \sqrt{9} = 5 - 3 = 2 \). Correct. Genuine. Both candidates survive, so the equation has two solutions. Why this one behaved well. At the intermediate step \( x + 1 = 4\sqrt{x - 2} \), the left side must be nonnegative for a solution to exist, requiring \( x \ge -1 \). Both candidates satisfy that comfortably, which is why neither was rejected on sign grounds. Contrast with the worked example. There the intermediate equation was \( \sqrt{2x + 5} = x + 1 \), and \( x = -2 \) made the right side negative. Checking the sign condition at each isolation step predicts which candidates will fail, though substituting into the original remains the definitive test. A verification of the algebra. The sum of the roots should be \( -\dfrac{b}{a} = 14 \), and \( 3 + 11 = 14 \). The product should be 33, and \( 3 \times 11 = 33 \). Both correct. \( x = 3 \) and \( x = 11 \), both genuine

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Give the domain of \( y = \sqrt{x - 7} \).
   Show the full solution

   \( x \ge 7 \)

2. Give the domain of \( y = \sqrt[3]{x - 7} \).
   Show the full solution

   Odd index, no restriction. All reals

3. Give the starting point of \( y = \sqrt{x + 3} \).
   Show the full solution

   Radicand zero at \( x = -3 \). \( (-3, 0) \)

4. Give the range of \( y = \sqrt{x} + 5 \).
   Show the full solution

   \( y \ge 5 \)

5. Does \( y = -\sqrt{x} \) open upward or downward from its start?
   Show the full solution

   Downward

6. Give the domain, range and starting point of \( y = 3\sqrt{2x - 6} + 1 \).
   Show the full solution

   Factor the radicand: \( 2x - 6 = 2(x - 3) \), so \( h = 3 \). Domain: \( 2x - 6 \ge 0 \), giving \( x \ge 3 \). Starting point: at \( x = 3 \) the radicand is 0, so \( y = 1 \). The point \( (3, 1) \). Range: \( a = 3 \) is positive, so \( y \ge 1 \). Check a second point: at \( x = 5 \) the radicand is 4, so \( y = 3(2) + 1 = 7 \). The point \( (5, 7) \). Domain \( x \ge 3 \), range \( y \ge 1 \), start \( (3, 1) \)

7. Give the domain and range of \( y = -\sqrt{5 - x} \).
   Show the full solution

   Radicand nonnegative: \( 5 - x \ge 0 \), so \( x \le 5 \). The domain runs to the left, because the variable has a negative coefficient inside the radical. That is a horizontal reflection. Starting point: at \( x = 5 \) the radicand is 0, so \( y = 0 \). The point \( (5, 0) \). Range: the outside negative makes every value nonpositive, so \( y \le 0 \). Check: at \( x = 1 \) the radicand is 4, so \( y = -2 \). The point \( (1, -2) \), which is left of the start and below the axis. Consistent. Domain \( x \le 5 \), range \( y \le 0 \)

8. Write the equation of \( y = \sqrt{x} \) reflected in the \( x \)-axis, shifted 2 left and 5 up, and give its range.
   Show the full solution

   Reflection in the \( x \)-axis: \( a = -1 \). Left 2: \( h = -2 \). Up 5: \( k = 5 \). \( y = -\sqrt{x + 2} + 5 \). Domain: \( x \ge -2 \). Starting point: \( (-2, 5) \). Range: the reflection sends the graph downward from the start, so \( y \le 5 \). Check at \( x = 2 \): radicand 4, so \( y = -2 + 5 = 3 \), which is below 5. Consistent. Check at \( x = 7 \): radicand 9, so \( y = -3 + 5 = 2 \). Still falling. Correct. \( y = -\sqrt{x+2} + 5 \), range \( y \le 5 \)

9. Explain why the graph of a square root function has a definite endpoint but a cube root function does not.
   Show the full solution

   The endpoint is where the domain stops, and the domain stops because of the even index. Square root. The radicand must be nonnegative, so there is a smallest permitted value of \( x \), the one making the radicand exactly zero. At that point the function is defined and takes the value \( k \), and for smaller \( x \) it is not defined at all. So the graph begins abruptly at \( (h, k) \) and does not continue leftward. Cube root. Every real number has a real cube root, so no value of \( x \) is forbidden. The point where the radicand is zero is not an endpoint but an ordinary interior point, and the graph passes through it and continues. What happens at that point instead. For the cube root, the point where the radicand vanishes is an inflection point: the curve flattens there and changes its direction of bending, but it does not stop or turn back. The shape consequence. A square root graph is half of a sideways parabola. A cube root graph is a full sideways cubic, stretching to infinity in both directions, because the cubic \( x = y^3 \) already passes the vertical line test and needs no branch removed. The general rule. Even index means a restricted domain and an endpoint; odd index means an unrestricted domain and no endpoint. It is the same parity distinction that governs absolute values in lesson 5.2 and extraneous roots in lesson 5.5. An even radicand must be nonnegative, creating a smallest permitted \( x \); an odd root accepts every real number

10. A pendulum's period in seconds is \( T = 2\pi\sqrt{\dfrac{L}{9.8}} \) with \( L \) in meters. Find the period of a 1 m pendulum, the length giving a 2-second period, and describe how the graph behaves.
    Show the full solution

    Find the period at \( L = 1 \). \( T = 2\pi\sqrt{\dfrac{1}{9.8}} = 2\pi\sqrt{0.10204} \). \( \sqrt{0.10204} \approx 0.31944 \). \( T \approx 2\pi(0.31944) \approx 6.28319 \times 0.31944 \approx 2.007 \) seconds. Find the length for a 2-second period. \( 2 = 2\pi\sqrt{\dfrac{L}{9.8}} \). Divide by \( 2\pi \): \( \sqrt{\dfrac{L}{9.8}} = \dfrac{1}{\pi} \approx 0.31831 \). Square both sides: \( \dfrac{L}{9.8} = \dfrac{1}{\pi^2} \approx 0.101321 \). \( L = 9.8 \times 0.101321 \approx 0.9929 \) meters. Check by substituting back. \( T = 2\pi\sqrt{\dfrac{0.9929}{9.8}} = 2\pi\sqrt{0.101317} \approx 2\pi(0.31831) = 2.000 \) seconds. Correct. Was squaring safe here? Yes. Both sides were positive before squaring, a period and a square root, so no sign information was lost and no extraneous root could appear. The check confirms it. Describe the graph. Writing \( T = \dfrac{2\pi}{\sqrt{9.8}}\sqrt{L} \approx 2.006\sqrt{L} \), it is the square root parent stretched vertically by about 2.006, with no shift. Domain: \( L \ge 0 \), and physically \( L \gt 0 \), since a pendulum of zero length is not a pendulum. Starting point: the origin, approached as the length shrinks. The behavior worth noticing. The graph rises steeply at first and then flattens. Going from 0.25 m to 1 m, a fourfold increase in length, only doubles the period, from about 1.00 s to about 2.01 s. Quadrupling the length doubles the period, because the relationship is a square root. Verify: \( T(0.25) \approx 2.006\sqrt{0.25} = 2.006(0.5) \approx 1.003 \) s. And \( T(4) \approx 2.006(2) = 4.01 \) s. So 1 m gives 2 s, 4 m gives 4 s. Confirmed. Why this matters historically. The near-coincidence that a 1 m pendulum has a period very close to 2 seconds, so that each swing takes about 1 second, made the pendulum the basis of accurate clocks for nearly three centuries. The flattening of the curve also means a small error in length causes an even smaller proportional error in period, which is what made such clocks reliable. What the model assumes. The formula holds for small swings; for large amplitudes the true period is longer, and the square root relationship is only an approximation. It also assumes no air resistance and a massless string. About 2.007 s for 1 m; about 0.993 m for a 2-second period

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Give the inverse of \( f(x) = x^3 \).
   Show the full solution

   \( \sqrt[3]{x} \)

2. Give the inverse of \( f(x) = x^2 \) restricted to \( x \ge 0 \).
   Show the full solution

   \( \sqrt{x} \)

3. Evaluate \( \sqrt{(-7)^2} \).
   Show the full solution

   \( \sqrt{49} \). 7

4. Evaluate \( \sqrt[3]{(-7)^3} \).
   Show the full solution

   Odd index preserves sign. \( -7 \)

5. Why does \( x^4 \) need a restricted domain to have an inverse?
   Show the full solution

   It is not one-to-one, since \( c \) and \( -c \) share an output

6. Find the inverse of \( f(x) = (x - 3)^3 + 1 \) and verify.
   Show the full solution

   Write \( y = (x - 3)^3 + 1 \) and swap: \( x = (y - 3)^3 + 1 \). \( x - 1 = (y - 3)^3 \). Cube root both sides: \( \sqrt[3]{x - 1} = y - 3 \). \( y = \sqrt[3]{x - 1} + 3 \). Verify: \( f(5) = 8 + 1 = 9 \), and \( f^{-1}(9) = \sqrt[3]{8} + 3 = 2 + 3 = 5 \). Round trip closed. Verify with a negative: \( f(1) = (-2)^3 + 1 = -7 \), and \( f^{-1}(-7) = \sqrt[3]{-8} + 3 = -2 + 3 = 1 \). Closed. No restriction is needed, since cubing is one-to-one. \( f^{-1}(x) = \sqrt[3]{x-1} + 3 \)

7. Find the inverse of \( f(x) = x^2 - 4 \) restricted to \( x \ge 0 \), and state its domain.
   Show the full solution

   Write \( y = x^2 - 4 \) and swap: \( x = y^2 - 4 \). \( y^2 = x + 4 \), so \( y = \pm\sqrt{x + 4} \). Choose the branch matching the restriction. The original was restricted to \( x \ge 0 \), so its outputs became the inverse's inputs and its inputs became the inverse's outputs. The inverse's range must therefore be \( y \ge 0 \), which selects the positive root: \( f^{-1}(x) = \sqrt{x + 4} \). Domain of the inverse. It is the range of the original. With \( x \ge 0 \), the smallest value of \( x^2 - 4 \) is \( -4 \), so the range is \( y \ge -4 \). The inverse's domain is \( x \ge -4 \), which also matches the radicand requirement. Verify: \( f(3) = 5 \), and \( f^{-1}(5) = \sqrt{9} = 3 \). Closed. \( f^{-1}(x) = \sqrt{x+4} \), domain \( x \ge -4 \)

8. Simplify \( \sqrt[4]{x^4} \) and \( \sqrt[5]{x^5} \), with no assumption on \( x \).
   Show the full solution

   Index 4 is even, so the principal root is nonnegative and the sign cannot be recovered: \( \sqrt[4]{x^4} = \left| x \right| \). Check at \( x = -2 \): \( \sqrt[4]{16} = 2 \), and \( \left| -2 \right| = 2 \). Correct. Writing \( x \) would have given \( -2 \), which is wrong. Index 5 is odd, so the root preserves sign: \( \sqrt[5]{x^5} = x \). Check at \( x = -2 \): \( \sqrt[5]{-32} = -2 \), since \( (-2)^5 = -32 \). And \( x = -2 \). Correct. The rule: even index gives \( \left| x \right| \), odd index gives \( x \). \( \left| x \right| \) and \( x \)

9. Explain why \( \left( \sqrt{x} \right)^2 \) and \( \sqrt{x^2} \) are different functions.
   Show the full solution

   Two functions are equal only when they have the same domain and the same values, and these differ on both counts. Domains. \( \left( \sqrt{x} \right)^2 \) requires \( \sqrt{x} \) to exist first, so its domain is \( x \ge 0 \). \( \sqrt{x^2} \) has radicand \( x^2 \), which is nonnegative for every real \( x \), so its domain is all reals. Already different. Values. On the shared domain \( x \ge 0 \), both give \( x \), so they agree there. On \( x \lt 0 \), the first is undefined and the second gives \( \left| x \right| \), a positive number. The order of operations explains it. In the first, rooting happens before squaring, so a negative input is rejected at the first step. In the second, squaring happens first and destroys the sign, so every input is accepted and the output carries no memory of the original sign. Graphically. \( \left( \sqrt{x} \right)^2 \) is the line \( y = x \) for \( x \ge 0 \) only, a ray from the origin. \( \sqrt{x^2} \) is the full absolute value graph, a V with its corner at the origin. They coincide on the right half and differ entirely on the left. Why this is the central fact of the unit. It is the statement that rooting and squaring undo each other in one order and not the other. Every restriction, absolute value and extraneous root in units 4 and 5 traces back to it. Different domains and different values: one is defined only for \( x \ge 0 \), the other is \( \left| x \right| \) on all reals

10. The volume of a sphere is \( V = \dfrac{4}{3}\pi r^3 \). Find the inverse function, state its domain, and find the radius of a sphere of volume 500 cubic cm.
    Show the full solution

    Solve for \( r \). \( V = \dfrac{4}{3}\pi r^3 \). Multiply both sides by \( \dfrac{3}{4\pi} \): \( r^3 = \dfrac{3V}{4\pi} \). Take the cube root: \( r = \sqrt[3]{\dfrac{3V}{4\pi}} \). State the domain. Mathematically the cube root accepts any real input, so the algebra imposes no restriction. Physically a volume cannot be negative, so the domain is \( V \ge 0 \), or \( V \gt 0 \) for a sphere that exists. This is the situation restricting the domain, not the algebra, which is the distinction lesson 4.7 emphasized. Had the formula involved a square root, the algebra would have restricted it too. Find the radius for \( V = 500 \). \( r = \sqrt[3]{\dfrac{3(500)}{4\pi}} = \sqrt[3]{\dfrac{1500}{12.56637}} = \sqrt[3]{119.3662} \). Estimating: \( 4.9^3 = 117.649 \) and \( 5.0^3 = 125 \). So \( r \) is between 4.9 and 5.0, closer to 4.9. \( 4.92^3 = 119.095 \) and \( 4.93^3 = 119.823 \). So \( r \approx 4.924 \) cm. Check by substituting back. \( V = \dfrac{4}{3}\pi(4.924)^3 = \dfrac{4}{3}\pi(119.389) \approx 1.33333 \times 3.14159 \times 119.389 \approx 500.1 \) cubic cm. Correct to rounding. Why no extraneous solution could appear. The inverse used a cube root, and cubing is one-to-one, so the operation is fully reversible. Each volume corresponds to exactly one radius, with no ambiguity to resolve. Contrast with a formula involving an even power. The area of a circle is \( A = \pi r^2 \), and inverting gives \( r = \pm\sqrt{\dfrac{A}{\pi}} \). Here the two branches genuinely exist mathematically, and the negative one is discarded because a radius cannot be negative. The cube root case needed no such choice. A sanity check on the size. A sphere of radius about 4.9 cm is roughly the size of a tennis ball, and 500 cubic cm is half a liter. Plausible. \( r = \sqrt[3]{\frac{3V}{4\pi}} \), domain \( V \gt 0 \); about 4.92 cm for 500 cubic cm

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Give the initial value of \( f(x) = 7 \cdot 3^x \).
   Show the full solution

   7

2. Give the growth factor of that function.
   Show the full solution

   3

3. Is \( f(x) = 4(0.5)^x \) growth or decay?
   Show the full solution

   The base is between 0 and 1. Decay

4. Give the horizontal asymptote of \( y = 2^x \).
   Show the full solution

   \( y = 0 \)

5. Evaluate \( 5 \cdot 2^3 \).
   Show the full solution

   Exponent before multiplication. 40

6. For \( f(x) = 6(1.5)^x \), find \( f(0) \), \( f(2) \) and \( f(-1) \).
   Show the full solution

   \( f(0) = 6 \). \( f(2) = 6(1.5)^2 = 6(2.25) = 13.5 \). \( f(-1) = 6(1.5)^{-1} = \dfrac{6}{1.5} = 4 \). Check the pattern: each step right multiplies by 1.5, and \( 4 \times 1.5 = 6 \), \( 6 \times 1.5 = 9 \), \( 9 \times 1.5 = 13.5 \). Consistent. 6, 13.5 and 4

7. Compare \( 2^x \) and \( x^2 \) at \( x = 2 \), \( x = 4 \) and \( x = 5 \).
   Show the full solution

   At \( x = 2 \): \( 2^2 = 4 \) and \( 2^2 = 4 \). Equal. At \( x = 4 \): \( 2^4 = 16 \) and \( 4^2 = 16 \). Equal again. At \( x = 5 \): \( 2^5 = 32 \) and \( 5^2 = 25 \). The exponential is ahead. They agree at exactly \( x = 2 \) and \( x = 4 \), and the exponential wins beyond that. Between 2 and 4 the square is larger: at \( x = 3 \), \( 2^3 = 8 \) and \( 3^2 = 9 \). Equal at 2 and 4; exponential larger at 5

8. A function has \( f(0) = 80 \) and multiplies by \( \dfrac{3}{4} \) each step. Write it and find \( f(3) \).
   Show the full solution

   Initial value 80, growth factor \( \dfrac{3}{4} = 0.75 \): \( f(x) = 80(0.75)^x \). \( f(3) = 80(0.75)^3 = 80(0.421875) = 33.75 \). Check step by step: \( 80 \to 60 \to 45 \to 33.75 \). Correct. Since the base is under 1, this is decay, losing 25 percent each step. \( f(x) = 80(0.75)^x \), \( f(3) = 33.75 \)

9. Explain why an exponential function with \( a \gt 0 \) never reaches zero.
   Show the full solution

   The value is \( ab^x \) with \( a \gt 0 \) and \( b \gt 0 \). A positive number raised to any real power is positive, never zero, so \( b^x \gt 0 \) always. Multiplying a positive by a positive gives a positive, so \( ab^x \gt 0 \) for every \( x \). What happens instead. For decay, the values shrink toward zero without reaching it. For \( 5(0.8)^x \), at \( x = 50 \) the value is about \( 5 \times 1.4 \times 10^{-5} \), tiny but positive. At \( x = 100 \) it is smaller still, and no finite \( x \) makes it zero. Why \( b^x \) cannot be zero. If \( b^x = 0 \) for some \( x \), then raising both sides to the power \( \dfrac{1}{x} \) would give \( b = 0 \), which is excluded. So no such \( x \) exists. The graphical statement. This is exactly what the horizontal asymptote \( y = 0 \) says: the curve approaches the axis arbitrarily closely and never touches it. The practical consequence. A decay model never predicts that a quantity reaches exactly zero. A drug is never fully cleared, a radioactive sample never fully decays. Real problems handle this by asking when the amount falls below some threshold rather than when it reaches zero. A positive base to any real power is positive, so the product with a positive \( a \) is always positive

10. Two functions pass through \( (0, 4) \). One is \( f(x) = 4 \cdot 3^x \) and the other is \( g(x) = 4 \cdot 3^{-x} \). Compare them and explain the relationship.
    Show the full solution

    Rewrite the second. Using \( 3^{-x} = \left( 3^{-1} \right)^x = \left( \dfrac{1}{3} \right)^x \), \( g(x) = 4\left( \dfrac{1}{3} \right)^x \). So \( g \) is an exponential with base \( \dfrac{1}{3} \), which is decay. Tabulate both. At \( x = -2 \): \( f = \dfrac{4}{9} \approx 0.444 \); \( g = 4(9) = 36 \). At \( x = -1 \): \( f = \dfrac{4}{3} \approx 1.333 \); \( g = 12 \). At \( x = 0 \): both 4. At \( x = 1 \): \( f = 12 \); \( g = \dfrac{4}{3} \approx 1.333 \). At \( x = 2 \): \( f = 36 \); \( g = \dfrac{4}{9} \approx 0.444 \). The relationship. The two tables are mirror images. Reading \( f \) left to right gives the same values as reading \( g \) right to left. Algebraically, \( g(x) = f(-x) \), which by lesson 1.3 is a reflection in the \( y \)-axis. Verify the reflection. \( f(-x) = 4 \cdot 3^{-x} = g(x) \). Confirmed directly. Both share the same asymptote. Reflecting in the \( y \)-axis leaves the horizontal asymptote \( y = 0 \) unchanged, since the reflection moves points horizontally and the asymptote is horizontal. Both approach zero, but from opposite sides: \( f \) as \( x \to -\infty \) and \( g \) as \( x \to \infty \). The general statement. Replacing \( b \) with \( \dfrac{1}{b} \) and replacing \( x \) with \( -x \) are the same operation. Every decay function can be written as a growth function with a negative exponent, and the two descriptions are interchangeable. That is why lesson 6.6 can convert freely between a half-life form and a base-\( e \) form. \( g \) is \( f \) reflected in the \( y \)-axis, since \( 3^{-x} = \left( \frac{1}{3} \right)^x \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Give the growth factor for a 5 percent increase.
   Show the full solution

   1.05

2. Give the decay factor for a 20 percent decrease.
   Show the full solution

   \( 1 - 0.20 \). 0.80

3. A table has outputs 4, 12, 36, 108. Find the common ratio.
   Show the full solution

   3

4. Write the model for that table if the first output is at \( x = 0 \).
   Show the full solution

   \( y = 4 \cdot 3^x \)

5. Write a model for 500 growing 8 percent per year.
   Show the full solution

   \( y = 500(1.08)^t \)

6. A table has outputs 64, 48, 36, 27 at \( x = 0, 1, 2, 3 \). Write the model.
   Show the full solution

   Ratios: \( \dfrac{48}{64} = 0.75 \), \( \dfrac{36}{48} = 0.75 \), \( \dfrac{27}{36} = 0.75 \). Constant. \( a = 64 \), \( b = 0.75 \). Model: \( y = 64(0.75)^x \). Check at \( x = 3 \): \( 64(0.421875) = 27 \). Correct. This is decay at 25 percent per step. \( y = 64(0.75)^x \)

7. An investment of $2000 grows 4.5 percent per year. Find its value after 12 years.
   Show the full solution

   Factor: \( 1.045 \). Model: \( A(t) = 2000(1.045)^t \). \( (1.045)^{12} \approx 1.695881 \). \( A(12) \approx 2000 \times 1.695881 \approx \$3391.76 \). Sanity check: simple interest would give \( 2000 + 12(90) = \$3080 \). Compounding adds about $312 more, which is the interest earned on previously earned interest. Plausible. About $3391.76

8. A substance halves every 3 days, starting at 48 grams. Write the model and find the amount after 12 days.
   Show the full solution

   The step is 3 days, so the exponent counts three-day periods: \( A(t) = 48\left( \dfrac{1}{2} \right)^{t/3} \). At \( t = 12 \): the exponent is \( \dfrac{12}{3} = 4 \), so \( A(12) = 48\left( \dfrac{1}{2} \right)^4 = 48 \times \dfrac{1}{16} = 3 \) grams. Check step by step: \( 48 \to 24 \to 12 \to 6 \to 3 \) over four three-day periods. Correct. \( A(t) = 48(0.5)^{t/3} \), giving 3 grams

9. Explain why the growth rate and the growth factor must not be confused.
   Show the full solution

   They measure different things. The rate is the fractional change, and the factor is what the quantity is multiplied by. For 6 percent growth, the rate is 0.06 and the factor is 1.06. The 1 accounts for keeping what was already there, and the 0.06 adds the growth on top. What goes wrong if they are swapped. Modeling with \( (0.06)^t \) describes a quantity keeping only 6 percent of itself each period, that is losing 94 percent. Starting at 1000, the correct model gives 1060 after one year while the wrong one gives 60. The check that catches it. Evaluate at \( t = 1 \) and compare with the description. If the answer should be slightly more than the start and it comes out far less, the factor is wrong. Where the confusion is most dangerous. With decay, both numbers are less than 1 and the error is less obvious. For a 15 percent loss, the factor is 0.85 and the rate is 0.15. Using 0.15 as the factor would describe an 85 percent loss per period, which after four periods gives \( 24000(0.15)^4 \approx \$12 \) instead of about $12,528. Three orders of magnitude apart. The rate is the fractional change; the factor is \( 1 \pm \) the rate, and it is the factor that goes in the model

10. A bacterial culture is measured at 50, 112, 245 and 560 cells at \( t = 0, 2, 4, 6 \) hours. Decide whether an exponential model fits, build one, and predict the count at \( t = 10 \).
    Show the full solution

    Test for a constant ratio. The inputs step by 2, so compare consecutive outputs: \( \dfrac{112}{50} = 2.24 \). \( \dfrac{245}{112} = 2.1875 \). \( \dfrac{560}{245} \approx 2.2857 \). Not exactly constant, but close. The three ratios range from 2.19 to 2.29, a spread of about 4 percent. Real measurements always scatter, so this is consistent with exponential growth plus counting error, and inconsistent with linear growth, whose differences would be constant instead. Check the linear alternative. Differences: \( 62 \), \( 133 \), \( 315 \). Nowhere near constant, and growing. Linear is clearly wrong. Build the model. Average the ratios: \( \dfrac{2.24 + 2.1875 + 2.2857}{3} \approx 2.2377 \) per two hours. With \( a = 50 \) and the exponent counting two-hour periods: \( N(t) = 50(2.2377)^{t/2} \). Convert to an hourly form. The hourly factor is \( \sqrt{2.2377} \approx 1.4959 \), so \( N(t) \approx 50(1.4959)^t \), a growth of about 49.6 percent per hour. Check the model against every data point. \( t = 2 \): \( 50(2.2377) \approx 111.9 \). Measured 112. Off by 0.1. \( t = 4 \): \( 50(2.2377)^2 = 50(5.0073) \approx 250.4 \). Measured 245. Off by about 2 percent. \( t = 6 \): \( 50(2.2377)^3 = 50(11.205) \approx 560.2 \). Measured 560. Off by 0.2. All within a few percent, which is good agreement for biological counts. Predict at \( t = 10 \). The exponent is \( \dfrac{10}{2} = 5 \). \( (2.2377)^5 = (2.2377)^3 \times (2.2377)^2 \approx 11.205 \times 5.0073 \approx 56.11 \). \( N(10) \approx 50 \times 56.11 \approx 2806 \) cells. How much to trust it. The prediction at \( t = 10 \) is an extrapolation four hours beyond the last measurement, which is a modest reach and probably reliable to within 10 percent. Where it stops being trustworthy. Exponential growth assumes unlimited nutrients and space. A real culture slows as the food runs out and eventually levels off, so the model overestimates increasingly badly at large \( t \). Extending it to \( t = 48 \) would predict \( 50(2.2377)^{24} \), a number around \( 10^{10} \), which no flask could hold. The honest statement is that the model describes the growth phase and says nothing about what follows. Exponential fits; \( N(t) \approx 50(1.4959)^t \), predicting about 2800 cells at 10 hours

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Round money to the nearest cent.

1. Give the periodic rate for 12 percent compounded monthly.
   Show the full solution

   \( \dfrac{0.12}{12} \). 0.01

2. Give the number of periods for 4 years compounded quarterly.
   Show the full solution

   \( 4 \times 4 \). 16

3. State the approximate value of \( e \).
   Show the full solution

   2.71828

4. Write the continuous compounding formula.
   Show the full solution

   \( A = Pe^{rt} \)

5. Find $500 at 4 percent compounded annually for 3 years.
   Show the full solution

   \( 500(1.04)^3 = 500(1.124864) \). About $562.43

6. Find $3000 at 5 percent compounded quarterly for 6 years.
   Show the full solution

   Periodic rate: \( \dfrac{0.05}{4} = 0.0125 \). Periods: \( 4 \times 6 = 24 \). \( A = 3000(1.0125)^{24} \). \( (1.0125)^{24} \approx 1.347351 \). \( A \approx 3000 \times 1.347351 \approx \$4042.05 \). Sanity check: simple interest would give \( 3000 + 6(150) = \$3900 \). Compounding adds about $142. Plausible. About $4042.05

7. Find $2000 at 7 percent compounded continuously for 10 years.
   Show the full solution

   \( A = 2000e^{0.07 \times 10} = 2000e^{0.7} \). \( e^{0.7} \approx 2.013753 \). \( A \approx 2000 \times 2.013753 \approx \$4027.51 \). Sanity check: the money has slightly more than doubled in 10 years at 7 percent, which matches the rule of thumb that 7 percent doubles in about 10 years. About $4027.51

8. Compare $5000 at 8 percent for 20 years compounded annually against continuously.
   Show the full solution

   Annually: \( 5000(1.08)^{20} \). \( (1.08)^{20} \approx 4.660957 \). \( A \approx \$23{,}304.79 \). Continuously: \( 5000e^{0.08 \times 20} = 5000e^{1.6} \). \( e^{1.6} \approx 4.953032 \). \( A \approx \$24{,}765.16 \). Difference: about $1460.37, roughly 6 percent more. Why the gap is larger here than in the worked example: both the rate and the time are larger, and the compounding advantage compounds too. Over 5 years at 6 percent the gap was under 1 percent; over 20 years at 8 percent it is over 6 percent. $23,304.79 against $24,765.16

9. Explain why \( e \) is called the natural base.
   Show the full solution

   The name records that \( e \) arises unavoidably from processes that grow in proportion to their own size, rather than being chosen for convenience. Where it comes from here. Continuous compounding means the interest is added at every instant, so the growth rate at any moment is proportional to the amount present. Solving that condition produces \( e^{rt} \) and nothing else. The number was not selected; it fell out. Why it recurs. The same condition describes radioactive decay, population growth without constraint, cooling toward room temperature, the discharge of a capacitor and the concentration of a drug in the bloodstream. Every one of them is modeled with base \( e \), because every one of them has a rate proportional to the current amount. The property that makes it special. In calculus, \( e^x \) is the only exponential function equal to its own rate of change. Any other base introduces a constant multiplier. That is the precise sense in which \( e \) is natural and 10 or 2 are not. Why other bases survive anyway. Base 2 is natural for doubling and for computing; base 10 matches decimal notation and the pH, Richter and decibel scales. All three are interchangeable through the change of base formula of lesson 7.4, so nothing is lost by preferring one. It arises automatically from any process growing in proportion to its own size, rather than being chosen

10. An account offers 5 percent compounded monthly and another offers 4.9 percent compounded continuously. Compare them over 15 years on a $10,000 deposit, and say which is better and why.
    Show the full solution

    First account: 5 percent monthly. Periodic rate: \( \dfrac{0.05}{12} \approx 0.00416667 \). Periods: \( 12 \times 15 = 180 \). \( A = 10000(1.00416667)^{180} \). \( (1.00416667)^{180} \approx 2.113704 \). \( A \approx \$21{,}137.04 \). Second account: 4.9 percent continuous. \( A = 10000e^{0.049 \times 15} = 10000e^{0.735} \). \( e^{0.735} \approx 2.085459 \). \( A \approx \$20{,}854.59 \). The first account wins by about $282.45 over 15 years. Why, when continuous compounding sounds better. Continuous compounding is the best possible use of a given rate, but it cannot make up an inferior rate. The 0.1 percentage point difference in rate outweighs the compounding advantage. Quantify the two effects separately. Compare 5 percent monthly against 5 percent continuous: \( 10000e^{0.75} = 10000(2.117000) \approx \$21{,}170.00 \). So continuous compounding at the same rate would add only about $32.96 over 15 years, or 0.16 percent. Meanwhile dropping the rate from 5 percent to 4.9 percent costs about $315, ten times as much. The effective annual rate settles it cleanly. For 5 percent monthly: \( (1.00416667)^{12} - 1 \approx 0.051162 \), that is 5.1162 percent. For 4.9 percent continuous: \( e^{0.049} - 1 \approx 0.050220 \), that is 5.0220 percent. The first has the higher effective rate, so it must win over any time period, and the 15-year comparison is just one instance of that. The general lesson. Compare accounts by effective annual rate, not by compounding frequency. Frequency is a second-order effect that advertising tends to emphasize precisely because the rate difference is unfavorable. The 5 percent monthly account, by about $282; its effective rate of 5.12 percent beats 5.02 percent

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Give the asymptote of \( y = 3^x + 5 \).
   Show the full solution

   \( y = 5 \)

2. Give the asymptote of \( y = 3^{\,x - 5} \).
   Show the full solution

   A horizontal shift does not move it. \( y = 0 \)

3. Give the \( y \)-intercept of \( y = 5 \cdot 2^x \).
   Show the full solution

   5

4. Give the range of \( y = 2^x - 4 \).
   Show the full solution

   \( y \gt -4 \)

5. Does \( y = -2^x \) lie above or below the \( x \)-axis?
   Show the full solution

   Below

6. For \( y = 3 \cdot 2^{\,x+1} - 6 \), give the asymptote, range and \( y \)-intercept.
   Show the full solution

   Comparing with \( a \cdot b^{x-h} + k \): \( x + 1 = x - (-1) \), so \( h = -1 \), and \( k = -6 \), \( a = 3 \). Asymptote: \( y = -6 \). Range: \( a \) positive, so \( y \gt -6 \). \( y \)-intercept: \( 3 \cdot 2^1 - 6 = 6 - 6 = 0 \). The graph passes through the origin. Check: at \( x = 1 \), \( y = 3(4) - 6 = 6 \). At \( x = -1 \), \( y = 3(1) - 6 = -3 \). Rising, and above \( -6 \). Correct. Asymptote \( y = -6 \), range \( y \gt -6 \), intercept 0

7. Write the equation of \( y = 2^x \) reflected in the \( x \)-axis and shifted up 8, and find its \( x \)-intercept.
   Show the full solution

   Reflection: \( a = -1 \). Up 8: \( k = 8 \). \( y = -2^x + 8 \). Asymptote: \( y = 8 \). Range: \( a \) negative, so \( y \lt 8 \). \( x \)-intercept: set \( -2^x + 8 = 0 \), so \( 2^x = 8 \) and \( x = 3 \). Check: \( -2^3 + 8 = -8 + 8 = 0 \). Correct. \( y \)-intercept: \( -1 + 8 = 7 \). \( y = -2^x + 8 \), \( x \)-intercept at \( x = 3 \)

8. A graph has asymptote \( y = 2 \), passes through \( (0, 5) \) and has base 3. Write its equation.
   Show the full solution

   The asymptote gives \( k = 2 \), so the form is \( y = a \cdot 3^x + 2 \), assuming no horizontal shift. Use the point: at \( x = 0 \), \( y = a(1) + 2 = a + 2 = 5 \), so \( a = 3 \). \( y = 3 \cdot 3^x + 2 \), which can also be written \( y = 3^{\,x+1} + 2 \). Check: at \( x = 0 \), \( 3 + 2 = 5 \). Correct. At \( x = 1 \), \( 9 + 2 = 11 \), and the alternative form gives \( 3^2 + 2 = 11 \). Agrees. The two forms are the same function, since a vertical stretch by the base is indistinguishable from a horizontal shift by 1 for an exponential. That is a property no other family has. \( y = 3 \cdot 3^x + 2 = 3^{x+1} + 2 \)

9. Explain why a vertical shift can give an exponential graph an \( x \)-intercept when the parent has none.
   Show the full solution

   The parent \( y = b^x \) has range \( y \gt 0 \), so it never takes the value zero and has no \( x \)-intercept. Its asymptote is the \( x \)-axis itself, approached but never met. What a downward shift does. Subtracting \( k \) moves the whole graph down, including its asymptote, which becomes \( y = -k \). The graph now lives above the line \( y = -k \) rather than above the axis, so part of it lies below the axis and it must cross. The condition. An \( x \)-intercept exists exactly when the asymptote and the graph lie on opposite sides of the axis. For \( y = a b^x + k \) with \( a \gt 0 \), that means \( k \lt 0 \). For \( a \lt 0 \) it means \( k \gt 0 \), as in problem 7. Finding it. Setting \( ab^x + k = 0 \) gives \( b^x = -\dfrac{k}{a} \), which has a solution exactly when the right side is positive. Solving for \( x \) generally needs a logarithm, which lesson 7.5 supplies; the cases in this lesson were arranged so the answer is a recognizable power. How many intercepts are possible. At most one, because an exponential is strictly increasing or strictly decreasing and therefore takes each value at most once. That contrasts with the quadratics of unit 2, which can have two. The parent's range excludes zero, but shifting moves the asymptote off the axis so the graph must cross it

10. A cup of coffee at \( 90^\circ \)C cools in a \( 20^\circ \)C room, losing 12 percent of the temperature difference each minute. Write a model, find the temperature after 10 minutes, and explain why the model has a vertical shift.
    Show the full solution

    Identify what decays. The coffee does not cool toward zero; it cools toward room temperature. What decays exponentially is the difference between the coffee and the room. Initial difference: \( 90 - 20 = 70 \) degrees. Model the difference. Losing 12 percent per minute means the decay factor is \( 1 - 0.12 = 0.88 \): \( D(t) = 70(0.88)^t \). Convert to temperature. The coffee's temperature is the room temperature plus the remaining difference: \( T(t) = 20 + 70(0.88)^t \). That is the vertical shift. The \( +20 \) raises the whole exponential so its asymptote is \( y = 20 \) rather than \( y = 0 \), which encodes the physical fact that the coffee approaches room temperature and not absolute zero. Evaluate at \( t = 10 \). \( (0.88)^{10} \): \( 0.88^2 = 0.7744 \); \( 0.88^4 = 0.7744^2 \approx 0.599695 \); \( 0.88^8 \approx 0.599695^2 \approx 0.359634 \); \( 0.88^{10} = 0.88^8 \times 0.88^2 \approx 0.359634 \times 0.7744 \approx 0.278502 \). \( T(10) \approx 20 + 70(0.278502) \approx 20 + 19.50 \approx 39.5^\circ \)C. Check the trajectory. \( T(0) = 20 + 70 = 90 \). Correct, the initial temperature. \( T(1) = 20 + 70(0.88) = 20 + 61.6 = 81.6 \). The coffee lost 8.4 degrees in the first minute, which is 12 percent of 70. Correct. \( T(5) = 20 + 70(0.88)^5 \). \( 0.88^5 = 0.88^4 \times 0.88 \approx 0.527732 \), so \( T(5) \approx 20 + 36.9 = 56.9^\circ \)C. Check the long-run behavior. \( T(30) \approx 20 + 70(0.88)^{30} \). Since \( 0.88^{30} = (0.88^{10})^3 \approx 0.278502^3 \approx 0.0216 \), \( T(30) \approx 21.5^\circ \)C. After half an hour the coffee is nearly at room temperature, as expected. Why the shift is essential. Without it, the model \( T(t) = 90(0.88)^t \) would predict the coffee cooling toward \( 0^\circ \)C, and at \( t = 30 \) it would give about \( 1.9^\circ \)C, which is below room temperature and physically impossible without a refrigerator. What this model is. It is Newton's law of cooling, which states that the rate of cooling is proportional to the temperature difference. Every application of it has this structure: an exponential decay of the difference, plus the ambient value as a vertical shift. \( T(t) = 20 + 70(0.88)^t \), giving about \( 39.5^\circ \)C after 10 minutes

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is constant for linear data?
   Show the full solution

   The first differences

2. What is constant for exponential data?
   Show the full solution

   The ratio of consecutive outputs

3. Classify 5, 10, 15, 20.
   Show the full solution

   Differences all 5. Linear

4. Classify 5, 10, 20, 40.
   Show the full solution

   Ratios all 2. Exponential

5. Does an exponential eventually exceed any polynomial?
   Show the full solution

   Yes

6. Classify 1, 4, 9, 16, 25 and give a formula.
   Show the full solution

   First differences: 3, 5, 7, 9. Not constant. Second differences: 2, 2, 2. Constant. Quadratic. The second difference is 2, so \( 2a = 2 \) and \( a = 1 \). Testing \( y = x^2 \) at \( x = 1 \) through 5 gives exactly the data. Quadratic, \( y = x^2 \)

7. Classify 100, 90, 81, 72.9 and give a formula.
   Show the full solution

   Differences: \( -10 \), \( -9 \), \( -8.1 \). Not constant. Ratios: \( \dfrac{90}{100} = 0.9 \), \( \dfrac{81}{90} = 0.9 \), \( \dfrac{72.9}{81} = 0.9 \). Constant. Exponential decay with \( b = 0.9 \). If the first value is at \( x = 0 \), the model is \( y = 100(0.9)^x \). Check at \( x = 3 \): \( 100(0.729) = 72.9 \). Correct. Exponential, \( y = 100(0.9)^x \)

8. Two savings plans: Plan A adds $2000 per year to an account starting at $10,000. Plan B starts at $10,000 and grows 8 percent per year. Compare after 5, 15 and 30 years.
   Show the full solution

   Plan A is linear: \( A(t) = 10000 + 2000t \). Plan B is exponential: \( B(t) = 10000(1.08)^t \). At \( t = 5 \). \( A = 10000 + 10000 = \$20{,}000 \). \( B = 10000(1.08)^5 = 10000(1.469328) \approx \$14{,}693 \). Plan A is well ahead. At \( t = 15 \). \( A = 10000 + 30000 = \$40{,}000 \). \( B = 10000(1.08)^{15} = 10000(3.172169) \approx \$31{,}722 \). Plan A still ahead, but the gap has narrowed in relative terms. At \( t = 30 \). \( A = 10000 + 60000 = \$70{,}000 \). \( B = 10000(1.08)^{30} = 10000(10.062657) \approx \$100{,}627 \). Plan B is now ahead by over $30,000. The crossover. Testing between: at \( t = 24 \), \( A = \$58{,}000 \) and \( B = 10000(6.341181) \approx \$63{,}412 \), so B is already ahead. At \( t = 22 \), \( A = \$54{,}000 \) and \( B = 10000(5.436540) \approx \$54{,}365 \), B barely ahead. At \( t = 21 \), \( A = \$52{,}000 \) and \( B \approx \$50{,}338 \), A ahead. So the crossover is between 21 and 22 years. Plan A leads for about 22 years, then Plan B pulls away permanently

9. Explain why a short table can make exponential and quadratic data look alike.
   Show the full solution

   Over a few steps both curve upward and both have growing increments, so the shapes are visually similar. The difference is in how the increments grow, and that takes several data points to become visible. A concrete near-collision. Compare \( y = 2^x \) and \( y = x^2 \) at \( x = 2, 3, 4 \): the values are 4, 8, 16 and 4, 9, 16. They agree at both ends and differ by only 1 in the middle. A three-point table there would not distinguish them. What separates them. Extending to \( x = 5 \) gives 32 and 25, and to \( x = 10 \) gives 1024 and 100. The gap becomes decisive, but only after the range is extended. The reliable test. Compute both the second differences and the ratios. For genuinely quadratic data the second differences are exactly constant; for exponential data the ratios are. Applying both tests to the same table settles it without needing more points. The judgment the test cannot make. Real data is never exact, so both tests give approximately constant results and the choice becomes a matter of which is closer. That is where the mechanism matters: ask whether the quantity grows by adding a fixed amount, by an amount that itself grows steadily, or by multiplying. A population reproducing multiplies; an object falling under gravity accumulates distance quadratically. The physical reasoning usually decides the model more reliably than the numbers do. Both curve upward with growing increments, and the difference in how the increments grow only becomes visible over a longer range

10. A rumor spreads so that the number of people who have heard it doubles every 3 hours, starting with 2 people. A town has 50,000 residents. Find when everyone has heard it, and explain what the model gets wrong.
    Show the full solution

    Build the model. Doubling every 3 hours from an initial 2: \( N(t) = 2 \cdot 2^{t/3} \), with \( t \) in hours. Tabulate by doublings. \( t = 0 \): 2. \( t = 3 \): 4. \( t = 6 \): 8. \( t = 15 \): \( 2 \cdot 2^5 = 64 \). \( t = 30 \): \( 2 \cdot 2^{10} = 2048 \). \( t = 39 \): \( 2 \cdot 2^{13} = 16{,}384 \). \( t = 42 \): \( 2 \cdot 2^{14} = 32{,}768 \). \( t = 45 \): \( 2 \cdot 2^{15} = 65{,}536 \). Find when it reaches 50,000. Between \( t = 42 \) and \( t = 45 \), since 32,768 is below and 65,536 is above. Testing \( t = 44 \): the exponent is \( \dfrac{44}{3} \approx 14.667 \), so \( N \approx 2 \cdot 2^{14.667} \). Since \( 2^{0.667} \approx 1.587 \), \( N \approx 2 \times 16384 \times 1.587 \approx 52{,}000 \). Just above. Testing \( t = 43.5 \): exponent 14.5, \( 2^{0.5} \approx 1.414 \), so \( N \approx 2 \times 16384 \times 1.414 \approx 46{,}340 \). Just below. So the model predicts everyone has heard it after roughly 43.8 hours, a little under two days. Check the scale of the answer. Going from 2 people to 50,000 requires about \( \log_2 25000 \approx 14.6 \) doublings, and at 3 hours each that is about 43.8 hours. Consistent. What the model gets wrong. First, it assumes every person who knows tells new people at a constant rate. In reality, as the rumor spreads, more and more of the people told already know it, so the effective rate falls. The growth slows long before saturation. Second, it treats the population as unlimited during the growth phase, but the last doubling would require 25,000 new listeners, which is half the town. There are not enough uninformed people left. Third, it produces non-integer counts, predicting 46,340.7 people at some instant, when the real quantity is a whole number. Fourth, it ignores that some people never hear rumors at all. What a better model looks like. Spread of this kind is usually modeled with a logistic curve, which grows exponentially at first, then bends over and levels off at the population size. The exponential model is the early portion of that curve and is accurate only while the informed fraction is small, here for perhaps the first 30 hours. Why the exponential model is still useful. It correctly captures the order of magnitude of the timescale, days rather than weeks or minutes, and it correctly identifies that the last few doublings happen astonishingly fast. Those two conclusions survive the model's flaws, and stating which conclusions survive is what makes a limited model worth reporting. About 44 hours; the model ignores saturation and overestimates the late stages

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A substance has a 10-year half-life. What fraction remains after 20 years?
   Show the full solution

   Two half-lives. \( \frac{1}{4} \)

2. What fraction remains after 30 years?
   Show the full solution

   Three half-lives. \( \frac{1}{8} \)

3. Write a model for 80 grams with a 6-day half-life.
   Show the full solution

   \( A(t) = 80(0.5)^{t/6} \)

4. A culture doubles every 4 hours from 25 cells. Find the count after 8 hours.
   Show the full solution

   Two doublings. 100 cells

5. Does a decaying quantity ever reach exactly zero?
   Show the full solution

   No

6. A 500 mg dose has a 6-hour half-life. Find the amount after 24 hours.
   Show the full solution

   \( \dfrac{24}{6} = 4 \) half-lives. \( A = 500\left( \dfrac{1}{2} \right)^4 = 500 \times \dfrac{1}{16} = 31.25 \) mg. Check: \( 500 \to 250 \to 125 \to 62.5 \to 31.25 \). Correct. 31.25 mg

7. A radioactive sample is at 12.5 percent of its original amount after 900 years. Find its half-life.
   Show the full solution

   12.5 percent is \( \dfrac{1}{8} = \left( \dfrac{1}{2} \right)^3 \), so three half-lives have passed. \( 3h = 900 \), giving \( h = 300 \) years. Check: after 300 years 50 percent remains, after 600 years 25 percent, after 900 years 12.5 percent. Correct. 300 years

8. A town of 8000 grows with a doubling time of 25 years. Write the model, find the population after 50 years, and give the annual growth rate.
   Show the full solution

   Model: \( P(t) = 8000 \cdot 2^{t/25} \). At \( t = 50 \): two doublings, so \( P = 8000 \times 4 = 32{,}000 \). Annual factor: \( 2^{1/25} \approx 1.028140 \), so about 2.81 percent per year. Check: \( (1.028140)^{25} \approx 2.0000 \), confirming the doubling. Cross-check with the rule of 70: \( \dfrac{70}{2.81} \approx 24.9 \) years, against the exact 25. Close. \( P(t) = 8000 \cdot 2^{t/25} \), 32,000 people, about 2.81 percent per year

9. Explain why the exponent is \( \dfrac{t}{h} \) rather than \( th \) or \( \dfrac{h}{t} \).
   Show the full solution

   The exponent must count how many half-lives have elapsed, because each half-life contributes one factor of \( \dfrac{1}{2} \). Elapsed time divided by the length of one period gives that count. With a half-life of 5 years, 15 years is \( \dfrac{15}{5} = 3 \) periods, so the exponent is 3 and the amount is \( \left( \dfrac{1}{2} \right)^3 \). Why \( th \) fails. It would give \( 15 \times 5 = 75 \) as the exponent, meaning 75 halvings after only 15 years. The prediction would be about \( 3 \times 10^{-23} \) of the original, which is absurd. Why \( \dfrac{h}{t} \) fails. It gives \( \dfrac{5}{15} = 0.333 \), meaning a third of one halving in 15 years, so the quantity would barely decrease. It also behaves backwards: more elapsed time would make the exponent smaller and the remaining amount larger. The one-line check. Substitute \( t = h \), one full period. The correct form gives an exponent of exactly 1 and therefore exactly half. Any other form fails this test, and it takes a few seconds. The units view. An exponent must be a pure number, not a quantity with units. Dividing years by years leaves a dimensionless count, while multiplying years by years would give square years, which cannot be an exponent. That check generalizes to every model of this kind. The exponent counts elapsed periods, which is time divided by the period's length; substituting \( t = h \) must give exactly one halving

10. A patient takes 300 mg of a drug with a 5-hour half-life every 10 hours. Find how much remains just before the second dose, just before the third, and explain what happens in the long run.
    Show the full solution

    Set up. Each dose decays according to \( A(t) = 300\left( \dfrac{1}{2} \right)^{t/5} \). Doses arrive at \( t = 0, 10, 20, \ldots \) hours. Just before the second dose, at \( t = 10 \). Only the first dose has been taken, and 10 hours is two half-lives. \( 300 \times \dfrac{1}{4} = 75 \) mg. The second dose is then added, bringing the total to \( 75 + 300 = 375 \) mg. Just before the third dose, at \( t = 20 \). Both previous doses have decayed. The first has been in the body 20 hours, four half-lives; the second has been in 10 hours, two half-lives. First dose remaining: \( 300 \times \dfrac{1}{16} = 18.75 \) mg. Second dose remaining: \( 300 \times \dfrac{1}{4} = 75 \) mg. Total: \( 93.75 \) mg. Check by a shortcut. The 375 mg present just after the second dose decays for 10 hours, two half-lives: \( 375 \times \dfrac{1}{4} = 93.75 \) mg. Agrees, and this is the faster route. Continue the pattern. After the third dose: \( 93.75 + 300 = 393.75 \) mg. Just before the fourth: \( 393.75 \times \dfrac{1}{4} = 98.44 \) mg. After the fourth: \( 398.44 \) mg. Just before the fifth: \( 99.61 \) mg. The long-run behavior. The trough values are climbing but by ever smaller amounts: 75, 93.75, 98.44, 99.61. They are approaching a limit. Find the limit. At steady state, the amount just before a dose, call it \( L \), must satisfy: add 300, then decay for two half-lives, and return to \( L \): \( (L + 300) \times \dfrac{1}{4} = L \). \( L + 300 = 4L \), so \( 3L = 300 \) and \( L = 100 \) mg. Verify. Starting at 100, adding 300 gives 400, and quartering gives 100. Stable. Correct. So the peaks approach 400 mg and the troughs approach 100 mg. The body reaches a steady state where each dose exactly replaces what was cleared. Why this matters clinically. The steady state is higher than a single dose, which is why a loading dose is sometimes given: to reach the therapeutic range immediately rather than after four or five doses. And the trough of 100 mg is the number that determines whether the drug stays effective between doses. The mathematics underneath. The trough values form a geometric series, and the limit found above is its sum. Lesson 8.6 develops infinite geometric series properly and would give the same 100 mg directly as \( \dfrac{300 \times \frac{1}{4}}{1 - \frac{1}{4}} = \dfrac{75}{0.75} = 100 \). 75 mg, then 93.75 mg; the troughs approach 100 mg and the peaks 400 mg

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does interpolation mean?
   Show the full solution

   Predicting inside the measured range

2. What does extrapolation mean?
   Show the full solution

   Predicting outside the measured range

3. Which is generally more reliable?
   Show the full solution

   Interpolation

4. Data has ratios 1.21, 1.19 and 1.20. Is exponential plausible?
   Show the full solution

   Nearly constant. Yes

5. Should a model be checked against one data point or all of them?
   Show the full solution

   All of them

6. Data at \( x = 0, 1, 2, 3 \) is 200, 240, 288, 346. Decide the model type and fit it.
   Show the full solution

   Differences: 40, 48, 58. Not constant, and growing. Ratios: \( \dfrac{240}{200} = 1.20 \), \( \dfrac{288}{240} = 1.20 \), \( \dfrac{346}{288} \approx 1.2014 \). Nearly constant. Exponential with \( b \approx 1.20 \) and \( a = 200 \). Model: \( y = 200(1.2)^x \). Check: \( x = 3 \) gives \( 200(1.728) = 345.6 \), against the measured 346. Off by 0.4, about 0.1 percent. Excellent. \( y = 200(1.2)^x \)

7. A model \( P(t) = 500(1.04)^t \) was fitted to data from \( t = 0 \) to \( t = 8 \). Evaluate at \( t = 6 \) and \( t = 40 \), and comment on each.
   Show the full solution

   At \( t = 6 \): \( (1.04)^6 \approx 1.265319 \), so \( P \approx 633 \). This is interpolation, inside the fitted range, and should be reliable to whatever accuracy the original fit achieved. At \( t = 40 \): \( (1.04)^{40} \approx 4.801021 \), so \( P \approx 2401 \). This is extrapolation 32 years beyond the data, four times the length of the fitted range. The arithmetic is correct but the prediction rests entirely on the assumption that 4 percent growth continues for four decades, which nothing in the data supports. 633 is a prediction; 2401 is a projection conditional on an unverified assumption

8. Two models are fitted to the same data. Model A misses by at most 2 percent; Model B misses by at most 0.5 percent but is a degree-6 polynomial. Which would you prefer and why?
   Show the full solution

   Model A, in most circumstances. Why the closer fit is not automatically better. A degree-6 polynomial has seven coefficients, enough freedom to pass very close to almost any small data set. Fitting the data well is therefore weak evidence that it has captured the underlying pattern; it may simply be tracing the noise. What that costs. A model that fits the noise behaves wildly outside the data range. High-degree polynomials typically swing off to large positive or negative values just beyond the last data point, so Model B's predictions would be unusable for extrapolation even though its fit looks better. When Model B would be preferable. If there is a reason to expect the underlying process to be a degree-6 polynomial, or if the only use is interpolation well inside the data range, the closer fit is worth having. The general principle. Prefer the simplest model consistent with the data, and prefer a model whose form matches the mechanism. A 2 percent miss from a model with two parameters is usually more informative than a 0.5 percent miss from one with seven. Model A, because a high-degree fit likely traces noise and behaves badly outside the data

9. Explain why a model should be reported with its range of validity.
   Show the full solution

   A formula does not know where it came from. It will return a number for any input, including inputs the data never covered and situations the model was never meant to describe. Reporting the range of validity is what prevents that number from being mistaken for a finding. What goes wrong without it. An exponential population model extended far enough predicts more people than the planet can hold. A linear model of a child's height extended far enough predicts a forty-foot adult. Both are arithmetically correct and both are worthless, and nothing inside the model flags the problem. What the range should be based on. The span of the data is the starting point: predictions inside it are interpolations and are supported. Beyond it, the question is whether the mechanism producing the pattern can reasonably be expected to continue, which is a judgment about the situation rather than about the numbers. What else belongs in the report. The largest disagreement between the model and the data, so a reader knows the accuracy. The assumptions, such as constant growth rate or unlimited resources. And what would make the model fail, which is often the most useful sentence in the whole analysis. Why this is a mathematical point and not just good manners. A model is a claim about a relationship, and a claim without its conditions is not a complete claim. Stating the domain is part of stating the function, exactly as it was in lesson 1.1. A formula returns a number for any input, so without a stated range nothing distinguishes a supported prediction from a meaningless one

10. Website visits are measured at 1200, 1560, 2028 and 2636 over four consecutive months. Fit a model, predict month 6, and assess how far the model can be pushed.
    Show the full solution

    Label the months. Take \( t = 0, 1, 2, 3 \) for the four measurements. Test linear. Differences: 360, 468, 608. Growing, not constant. Not linear. Test exponential. Ratios: \( \dfrac{1560}{1200} = 1.30 \). \( \dfrac{2028}{1560} = 1.30 \). \( \dfrac{2636}{2028} \approx 1.29980 \). Constant to four decimal places. Strongly exponential. Fit the model. \( V(t) = 1200(1.3)^t \), with \( t \) in months. Check against every point. \( t = 1 \): \( 1200(1.3) = 1560 \). Exact. \( t = 2 \): \( 1200(1.69) = 2028 \). Exact. \( t = 3 \): \( 1200(2.197) = 2636.4 \). Measured 2636. Off by 0.4, under 0.02 percent. The fit is essentially perfect, which suggests the data was generated by a clean process or has been rounded. Predict month 6, that is \( t = 5 \). \( (1.3)^5 = (1.3)^2 \times (1.3)^3 = 1.69 \times 2.197 = 3.71293 \). \( V(5) \approx 1200 \times 3.71293 \approx 4456 \) visits. Check the intermediate month. \( t = 4 \) gives \( 1200(2.8561) \approx 3427 \), so the sequence runs 2636, 3427, 4456. The growth is accelerating in absolute terms, as exponential growth does. How far can this be pushed? Short reach is reasonable. Two months beyond four months of data is a modest extrapolation, and the fit is exceptionally tight, so 4456 is a defensible projection. Longer reach is not. At \( t = 12 \), one year from the start, \( 1200(1.3)^{12} \approx 1200 \times 23.30 \approx 27{,}960 \) visits. At \( t = 24 \), \( 1200(1.3)^{24} \approx 1200 \times 542.8 \approx 651{,}000 \). At \( t = 36 \), about 15.2 million. At \( t = 48 \), about 353 million, which exceeds the population of the United States. Why it must break down. Thirty percent monthly growth means roughly a 23-fold increase per year. No website sustains that for long, because the pool of potential visitors is finite and marketing, server capacity and interest all saturate. What a better long-run model would look like. A logistic curve, growing exponentially at first and leveling off at some ceiling set by the addressable audience. The exponential model is the early part of that curve. The recommendation. Use \( V(t) = 1200(1.3)^t \) for projections up to about month 8, report the month 6 figure of roughly 4450 visits, and collect more data before projecting further. State that the 30 percent monthly rate is the assumption the whole projection depends on and that it is the first thing to re-examine. \( V(t) = 1200(1.3)^t \), about 4456 visits in month 6; trustworthy for a few months and not beyond

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Evaluate \( \log_3 9 \).
   Show the full solution

   \( 3^2 = 9 \). 2

2. Evaluate \( \log_5 1 \).
   Show the full solution

   Any base to the zero power is 1. 0

3. Evaluate \( \log_{10} 10000 \).
   Show the full solution

   \( 10^4 \). 4

4. Write \( 2^6 = 64 \) in logarithmic form.
   Show the full solution

   \( \log_2 64 = 6 \)

5. Is \( \log(-8) \) defined over the reals?
   Show the full solution

   No power of 10 is negative. No

6. Evaluate \( \log_4 \dfrac{1}{64} \).
   Show the full solution

   \( 64 = 4^3 \), so \( \dfrac{1}{64} = 4^{-3} \). \( -3 \)

7. Evaluate \( \log_9 27 \).
   Show the full solution

   Write both in base 3: \( 9 = 3^2 \) and \( 27 = 3^3 \). Let \( \log_9 27 = y \), so \( 9^y = 27 \), that is \( 3^{2y} = 3^3 \). Equating exponents: \( 2y = 3 \), so \( y = \dfrac{3}{2} \). Check: \( 9^{3/2} = \left( \sqrt{9} \right)^3 = 27 \). Correct. A logarithm need not be a whole number, and rewriting both base and argument as powers of a common base is the way to find an exact fractional value. \( \frac{3}{2} \)

8. Solve \( \log_x 49 = 2 \) for \( x \).
   Show the full solution

   Convert to exponential form: \( x^2 = 49 \), so \( x = \pm 7 \). But a base must be positive and not 1, so \( x = -7 \) is rejected. \( x = 7 \). Check: \( \log_7 49 = 2 \) since \( 7^2 = 49 \). Correct. \( x = 7 \)

9. Explain why a logarithm can be negative even though its argument cannot.
   Show the full solution

   The two are different quantities with different restrictions, and the definition makes the difference visible. The argument is the output of an exponential. In \( \log_b x = y \), the \( x \) is \( b^y \), and a positive base raised to any power is positive. So the argument is confined to positive values. The logarithm is the exponent. An exponent may be any real number: positive, zero or negative. A negative exponent produces a value between 0 and 1, which is still positive. Concretely. \( \log_2 \dfrac{1}{8} = -3 \). The argument \( \dfrac{1}{8} \) is positive and the logarithm \( -3 \) is negative, and both are legitimate. The pattern to remember. A logarithm is negative exactly when its argument is between 0 and 1, zero when the argument is 1, and positive when the argument exceeds 1. That reads directly off the graph in lesson 7.2. The argument is a positive power of the base; the logarithm is the exponent, and exponents may be negative

10. Evaluate \( \log_2 \left( \log_3 \left( \log_5 125 \right) \right) \).
    Show the full solution

    Work from the inside out. Innermost. \( \log_5 125 \) asks what power of 5 gives 125. Since \( 5^3 = 125 \), the value is 3. Middle. The expression becomes \( \log_3 3 \), which asks what power of 3 gives 3. The answer is 1. Outermost. The expression becomes \( \log_2 1 \), which asks what power of 2 gives 1. Since \( 2^0 = 1 \), the answer is 0. 0 Check every stage is legal. Each argument must be positive: 125 is, 3 is, and 1 is. So no step is undefined. A caution about what comes next. Had the innermost evaluated to 1, the middle would give \( \log_3 1 = 0 \), and the outermost would be \( \log_2 0 \), which is undefined. Nested logarithms can fail even when each individual base and argument looks reasonable, so the domain has to be checked at every level rather than only at the start. Why the answer is exactly 0 and not approximately. Every step used an exact power relationship, so no rounding entered anywhere. Evaluating with a calculator would give 0 as well, but the exact route shows why.

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Give the domain of \( y = \log_5 x \).
   Show the full solution

   \( x \gt 0 \)

2. Give the vertical asymptote of \( y = \log x \).
   Show the full solution

   \( x = 0 \)

3. Through which point does every graph \( y = \log_b x \) pass?
   Show the full solution

   \( (1, 0) \)

4. Give the range of \( y = \log_2 x \).
   Show the full solution

   All real numbers

5. In which line is \( y = \log_b x \) the reflection of \( y = b^x \)?
   Show the full solution

   \( y = x \)

6. Give the domain and asymptote of \( y = \log_3(x + 4) \).
   Show the full solution

   The argument must be positive: \( x + 4 \gt 0 \), so \( x \gt -4 \). Asymptote: \( x = -4 \). Check a point: at \( x = -1 \) the argument is 3, so \( y = 1 \), giving \( (-1, 1) \). The shift is left 4, since \( x + 4 = x - (-4) \). Domain \( x \gt -4 \), asymptote \( x = -4 \)

7. Give the domain and asymptote of \( y = \log_2(6 - x) \).
   Show the full solution

   Argument positive: \( 6 - x \gt 0 \), so \( x \lt 6 \). The domain runs to the left, because the variable has a negative coefficient inside, which is a horizontal reflection. Asymptote: \( x = 6 \). Check: at \( x = 2 \) the argument is 4, so \( y = 2 \). At \( x = 5 \) the argument is 1, so \( y = 0 \). The graph falls as \( x \) increases toward 6. Domain \( x \lt 6 \), asymptote \( x = 6 \)

8. Write the equation of \( y = \log_2 x \) shifted 1 right and 3 down, and give its \( x \)-intercept.
   Show the full solution

   \( y = \log_2(x - 1) - 3 \). Domain: \( x \gt 1 \). Asymptote: \( x = 1 \). \( x \)-intercept: set \( y = 0 \): \( \log_2(x - 1) = 3 \), so \( x - 1 = 2^3 = 8 \) and \( x = 9 \). Check: \( \log_2(8) - 3 = 3 - 3 = 0 \). Correct. \( y = \log_2(x-1) - 3 \), \( x \)-intercept at \( x = 9 \)

9. Explain why a logarithmic graph has a vertical asymptote rather than a horizontal one.
   Show the full solution

   It inherits it from the exponential by reflection. The exponential's asymptote. \( y = b^x \) approaches \( y = 0 \) as \( x \) runs to negative infinity, so it has a horizontal asymptote along the \( x \)-axis. What reflection does to it. Reflecting in \( y = x \) swaps the two axes. A horizontal line through the origin becomes a vertical line through the origin, so the asymptote \( y = 0 \) becomes \( x = 0 \). The direct reason, without reflection. As \( x \) approaches 0 from the right, the question "what power of \( b \) gives \( x \)?" requires an increasingly large negative exponent. To get \( 2^y = 0.001 \) needs \( y \approx -9.97 \); to get \( 2^y = 0.000001 \) needs \( y \approx -19.9 \). The output falls without bound while the input stays near zero, which is exactly what a vertical asymptote is. Why it never crosses. The argument cannot be zero or negative, so the graph has no points at or left of \( x = 0 \). The asymptote is not merely approached; it is the edge of the domain. Reflection in \( y = x \) turns the exponential's horizontal asymptote into a vertical one

10. The graphs of \( y = 2^x \) and \( y = \log_2 x \) both approach the line \( y = x \) without crossing it. Verify this at three points and explain what it means.
    Show the full solution

    Check \( y = 2^x \) against \( y = x \). At \( x = 1 \): \( 2^1 = 2 \), which exceeds 1. At \( x = 0.5 \): \( 2^{0.5} \approx 1.414 \), which exceeds 0.5. At \( x = 0 \): \( 2^0 = 1 \), which exceeds 0. At \( x = -1 \): \( 2^{-1} = 0.5 \), which exceeds \( -1 \). The exponential lies above \( y = x \) everywhere. Find where they come closest. The gap \( 2^x - x \) is 1 at \( x = 0 \), about 0.914 at \( x = 0.5 \), 1 at \( x = 1 \). Testing between: at \( x = 0.5 \) the gap is 0.914; at \( x = 0.6 \), \( 2^{0.6} \approx 1.516 \), gap 0.916. So the minimum gap is near \( x = 0.53 \) and is about 0.914. It never reaches zero. Check \( y = \log_2 x \) against \( y = x \). At \( x = 1 \): \( \log_2 1 = 0 \), which is below 1. At \( x = 2 \): \( \log_2 2 = 1 \), below 2. At \( x = 0.5 \): \( \log_2 0.5 = -1 \), below 0.5. The logarithm lies below \( y = x \) everywhere. Why the two facts are the same fact. The two graphs are reflections of each other in \( y = x \). If one lies entirely above that line, its mirror image must lie entirely below it. So verifying either statement establishes the other. What it means about solutions. The equation \( 2^x = x \) has no real solution, because the graphs never meet. Neither does \( \log_2 x = x \). That is worth knowing because such equations look solvable and are not. The contrast with a smaller base. This depends on the base being large enough. For \( y = 1.2^x \), the curve does cross \( y = x \): at \( x = 1 \) it gives 1.2, above the line, but at \( x = 2 \) it gives 1.44, below 2. So a crossing occurs between 1 and 2. The general statement holds for bases above \( e^{1/e} \approx 1.4447 \), which is a genuinely surprising threshold and one worth knowing exists even before Precalculus explains it. Both verified; the exponential stays above \( y = x \) and the logarithm below, because they are reflections in that line

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Expand \( \log(xy) \).
   Show the full solution

   \( \log x + \log y \)

2. Expand \( \log \dfrac{x}{y} \).
   Show the full solution

   \( \log x - \log y \)

3. Expand \( \log(x^5) \).
   Show the full solution

   \( 5\log x \)

4. Condense \( \log 3 + \log 4 \).
   Show the full solution

   \( \log 12 \). \( \log 12 \)

5. Does \( \log(x + y) \) simplify?
   Show the full solution

   No

6. Expand \( \ln \dfrac{\sqrt{x}}{y^3} \) completely.
   Show the full solution

   Quotient first: \( \ln\sqrt{x} - \ln(y^3) \). Write the radical as an exponent: \( \sqrt{x} = x^{1/2} \). Power property on each: \( \dfrac{1}{2}\ln x - 3\ln y \). Converting the radical to a rational exponent is what lets the power property apply, which is lesson 5.1 earning its place here. \( \frac{1}{2}\ln x - 3\ln y \)

7. Condense \( 3\log_2 x - \dfrac{1}{2}\log_2 y + \log_2 5 \).
   Show the full solution

   Move the coefficients inside as exponents: \( \log_2(x^3) - \log_2(y^{1/2}) + \log_2 5 \). Combine: addition means multiply, subtraction means divide: \( \log_2 \dfrac{5x^3}{\sqrt{y}} \). Check by expanding back: \( \log_2(5x^3) - \log_2(y^{1/2}) = \log_2 5 + 3\log_2 x - \frac{1}{2}\log_2 y \). Matches. \( \log_2 \frac{5x^3}{\sqrt{y}} \)

8. Given \( \log 2 \approx 0.3010 \) and \( \log 3 \approx 0.4771 \), evaluate \( \log 12 \) and \( \log 1.5 \) without a calculator.
   Show the full solution

   \( \log 12 \). Factor: \( 12 = 2^2 \times 3 \). \( \log 12 = 2\log 2 + \log 3 \approx 2(0.3010) + 0.4771 = 0.6020 + 0.4771 = 1.0791 \). \( \log 1.5 \). Write it as a quotient: \( 1.5 = \dfrac{3}{2} \). \( \log 1.5 = \log 3 - \log 2 \approx 0.4771 - 0.3010 = 0.1761 \). Check both. \( 10^{1.0791} \approx 12.0 \) and \( 10^{0.1761} \approx 1.50 \). Correct. This is what logarithm tables were for before calculators: knowing a handful of values let every product, quotient and power be computed by addition. About 1.0791 and 0.1761

9. Explain why the logarithm properties are just the exponent rules restated.
   Show the full solution

   A logarithm is an exponent, so any statement about how exponents combine becomes a statement about how logarithms combine. Property by property. \( b^x b^y = b^{x+y} \) says multiplying powers adds exponents. Reading it through the definition, with \( M = b^x \) and \( N = b^y \), it says the logarithm of a product is the sum of the logarithms. That is the product property. \( \dfrac{b^x}{b^y} = b^{x-y} \) becomes the quotient property by the same substitution. \( (b^x)^p = b^{xp} \) becomes the power property: raising to a power multiplies the exponent, so it multiplies the logarithm. Why there is no sum property. There is no exponent rule for \( b^x + b^y \). Adding powers of the same base does not simplify, so neither does the logarithm of a sum. The absence is not an oversight; it mirrors an absence one level down. What this buys practically. A forgotten logarithm property can be rederived in two lines from the corresponding exponent rule. And the pattern predicts which manipulations are legal without memorizing a list. A logarithm is an exponent, so each property is an exponent rule read through the definition; there is no sum rule because there is none for exponents either

10. A student writes \( \dfrac{\log 8}{\log 2} = \log 4 \). Show this is wrong, find the correct value, and explain what property was misapplied.
    Show the full solution

    Evaluate both sides. \( \log 8 \approx 0.9031 \) and \( \log 2 \approx 0.3010 \), so the left side is \( \dfrac{0.9031}{0.3010} \approx 3.000 \). \( \log 4 \approx 0.6021 \). These are not equal, so the claim is false. The correct value is exactly 3. By the change of base formula of lesson 7.4, \( \dfrac{\log 8}{\log 2} = \log_2 8 = 3 \), since \( 2^3 = 8 \). The decimal 3.000 confirms it. What was misapplied. The student used the quotient property backwards. That property says \( \log M - \log N = \log \dfrac{M}{N} \): a difference of logarithms becomes the logarithm of a quotient. A quotient of logarithms is something entirely different and does not simplify that way. The correct difference, for contrast. \( \log 8 - \log 2 = \log 4 \approx 0.6021 \), and checking, \( 0.9031 - 0.3010 = 0.6021 \). Correct. So the student's answer is right for the wrong expression. The pair worth keeping apart. \( \log M - \log N = \log \dfrac{M}{N} \), a difference. \( \dfrac{\log M}{\log N} = \log_N M \), a quotient, which is the change of base formula. Both are true and they say completely different things. A memorable check. \( \dfrac{\log 100}{\log 10} = \dfrac{2}{1} = 2 \), which is \( \log_{10} 100 \). Meanwhile \( \log 100 - \log 10 = 2 - 1 = 1 \), which is \( \log 10 \). Two different operations, two different answers. The value is 3; the quotient property applies to a difference of logarithms, not a quotient of them

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Round to four decimal places.

1. Write \( \log_5 30 \) using natural logarithms.
   Show the full solution

   \( \frac{\ln 30}{\ln 5} \)

2. Which goes on top, the base or the argument?
   Show the full solution

   The argument

3. Evaluate \( \log_2 10 \).
   Show the full solution

   \( \dfrac{\ln 10}{\ln 2} \approx \dfrac{2.3026}{0.6931} \). About 3.3219

4. Between which whole numbers does \( \log_4 100 \) lie?
   Show the full solution

   \( 4^3 = 64 \) and \( 4^4 = 256 \). Between 3 and 4

5. Evaluate \( \log_4 100 \).
   Show the full solution

   \( \dfrac{\ln 100}{\ln 4} \approx \dfrac{4.6052}{1.3863} \). About 3.3219

6. Evaluate \( \log_6 250 \), estimating first.
   Show the full solution

   Estimate: \( 6^3 = 216 \) and \( 6^4 = 1296 \), so the answer is just above 3. \( \log_6 250 = \dfrac{\ln 250}{\ln 6} \approx \dfrac{5.5215}{1.7918} \approx 3.0816 \). Just above 3, as predicted. Check: \( 6^{3.0816} = e^{3.0816 \times 1.7918} = e^{5.5216} \approx 250.0 \). Correct. About 3.0816

7. Problems 3 and 5 gave the same answer. Explain why.
   Show the full solution

   Because \( 4 = 2^2 \) and \( 100 = 10^2 \), so both logarithms are asking the same proportional question. Algebraically. \( \log_4 100 = \dfrac{\ln 100}{\ln 4} = \dfrac{\ln(10^2)}{\ln(2^2)} = \dfrac{2\ln 10}{2\ln 2} = \dfrac{\ln 10}{\ln 2} = \log_2 10 \). The factor of 2 appears in both the numerator and the denominator and cancels. The general fact. \( \log_{b^n}(x^n) = \log_b x \) for any \( n \). Raising the base and the argument to the same power leaves the logarithm unchanged. Check with another instance. \( \log_8 1000 = \dfrac{\ln 1000}{\ln 8} = \dfrac{3\ln 10}{3\ln 2} = \log_2 10 \approx 3.3219 \). The same value again. Raising base and argument to the same power cancels out

8. Rewrite \( f(x) = \log_5 x \) in terms of \( \ln \), and state the transformation this represents.
   Show the full solution

   \( \log_5 x = \dfrac{\ln x}{\ln 5} = \dfrac{1}{\ln 5}\ln x \). Since \( \ln 5 \approx 1.6094 \), the coefficient is \( \dfrac{1}{1.6094} \approx 0.6213 \). So \( f(x) \approx 0.6213 \ln x \). The transformation is a vertical compression of \( \ln x \) by a factor of about 0.62, since the coefficient is between 0 and 1. Check at \( x = 5 \): \( \log_5 5 = 1 \), and \( 0.6213 \ln 5 = 0.6213(1.6094) = 1.0000 \). Correct. Why every logarithmic graph has the same shape. Changing the base only scales the graph vertically. All logarithmic functions are vertical stretches or compressions of one another, which is why they share the asymptote, the point \( (1, 0) \) and the overall form. \( f(x) = \frac{\ln x}{\ln 5} \), a vertical compression by about 0.62

9. Explain why the formula cannot use a new base of 1.
   Show the full solution

   The formula divides by \( \log_a b \), and with \( a = 1 \) that quantity is not defined at all. Why \( \log_1 b \) fails. It asks what power of 1 gives \( b \). But \( 1^y = 1 \) for every \( y \), so the question has no answer when \( b \ne 1 \), and infinitely many answers when \( b = 1 \). Neither case defines a function value. That is exactly why base 1 was excluded in lesson 7.1. The restriction is not arbitrary; a base of 1 makes the exponential constant, so it is not one-to-one and has no inverse. The parallel restriction on the original base. The same argument excludes \( b = 1 \) from \( \log_b x \) itself. And negative bases are excluded because fractional powers of a negative are not real. What remains. Any base that is positive and not 1 works, and all such choices give the same answer, as the worked example showed. Base 10 and base \( e \) are conventional only because calculators provide them. \( \log_1 b \) is undefined, since every power of 1 is 1, so the formula's denominator would not exist

10. A calculator shows only \( \log \) and \( \ln \). Use it to find the exact value of \( \log_8 32 \), and verify the exact answer independently.
    Show the full solution

    Compute with the calculator. \( \log_8 32 = \dfrac{\ln 32}{\ln 8} \approx \dfrac{3.4657}{2.0794} \approx 1.6667 \). The decimal repeats, suggesting an exact fraction of \( \dfrac{5}{3} \). Confirm exactly, without the calculator. Write both base and argument as powers of 2: \( 8 = 2^3 \) and \( 32 = 2^5 \). Let \( y = \log_8 32 \), so \( 8^y = 32 \), that is \( 2^{3y} = 2^5 \). Equating exponents: \( 3y = 5 \), so \( y = \dfrac{5}{3} \) exactly. Check. \( 8^{5/3} = \left( \sqrt[3]{8} \right)^5 = 2^5 = 32 \). Correct. Confirm the decimal matches. \( \dfrac{5}{3} = 1.6\overline{6} \), and the calculator gave 1.6667. Agrees. The general shortcut this illustrates. When the base and the argument are powers of a common number, the logarithm is the ratio of the exponents: \( \log_{a^m}(a^n) = \dfrac{n}{m} \). Here \( \dfrac{5}{3} \). Why the exact route is worth taking. The calculator gives a decimal that might be \( \dfrac{5}{3} \) or might be something near it. Only the algebra establishes that it is exactly \( \dfrac{5}{3} \), and an exact answer is what a question asking for one requires. A caution. The shortcut works only when a common base exists. \( \log_8 30 \) has no such simplification, because 30 is not a power of 2, and the change of base formula with a decimal answer is then the best available. Exactly \( \frac{5}{3} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( 2^x = 32 \).
   Show the full solution

   \( 32 = 2^5 \). \( x = 5 \)

2. Solve \( 3^x = 27 \).
   Show the full solution

   \( x = 3 \)

3. Solve \( 10^x = 1000 \).
   Show the full solution

   \( x = 3 \)

4. Solve \( 4^{x} = 64 \).
   Show the full solution

   \( 64 = 4^3 \). \( x = 3 \)

5. Solve \( e^x = 10 \), exactly.
   Show the full solution

   Take \( \ln \) of both sides. \( x = \ln 10 \approx 2.3026 \)

6. Solve \( 2^{x-3} = 16 \).
   Show the full solution

   \( 16 = 2^4 \), so \( 2^{x-3} = 2^4 \). Equating exponents: \( x - 3 = 4 \), so \( x = 7 \). Check: \( 2^{7-3} = 2^4 = 16 \). Correct. \( x = 7 \)

7. Solve \( 4 \cdot 3^x = 36 \).
   Show the full solution

   Isolate first. Divide both sides by 4: \( 3^x = 9 \). Then \( 9 = 3^2 \), so \( x = 2 \). Check: \( 4 \cdot 3^2 = 4(9) = 36 \). Correct. Isolating turned a problem that looked as if it needed logarithms into one that did not. \( x = 2 \)

8. Solve \( 7^{2x} = 100 \).
   Show the full solution

   Take natural logarithms: \( 2x\ln 7 = \ln 100 \). \( x = \dfrac{\ln 100}{2\ln 7} \approx \dfrac{4.6052}{2(1.9459)} = \dfrac{4.6052}{3.8918} \approx 1.1833 \). Estimate check: \( 7^2 = 49 \) and \( 7^{2.5} \approx 129 \), so \( 2x \) is between 2 and 2.5, making \( x \) between 1 and 1.25. Consistent. Check: \( 7^{2(1.1833)} = 7^{2.3666} = e^{2.3666 \times 1.9459} = e^{4.6051} \approx 100.0 \). Correct. About 1.1833

9. Explain why matching bases and taking logarithms give the same answer.
   Show the full solution

   They are the same operation, with one route hiding the logarithm inside a recognizable power. Matching bases. From \( b^{f(x)} = b^{g(x)} \), concluding \( f(x) = g(x) \) uses that the exponential is one-to-one. Taking logarithms. Applying \( \log_b \) to both sides gives \( \log_b(b^{f(x)}) = \log_b(b^{g(x)}) \), and since \( \log_b b^u = u \) by lesson 7.1, this reduces to \( f(x) = g(x) \). Exactly the same conclusion. So matching bases is taking a base-\( b \) logarithm, done mentally because the result is obvious. A demonstration. Solve \( 2^x = 8 \) both ways. Matching: \( 8 = 2^3 \), so \( x = 3 \). By logarithms: \( x = \dfrac{\ln 8}{\ln 2} = \dfrac{2.0794}{0.6931} = 3.000 \). Identical. Which to use. Match bases when it is quick, because the answer is exact and requires no calculator. Use logarithms otherwise. The logarithm route always works; the matching route only sometimes does. Matching bases is applying a base-\( b \) logarithm and simplifying in one step

10. A radioactive sample decays with a half-life of 12 years. Find how long until 10 percent remains, and check the answer against the half-life directly.
    Show the full solution

    Write the model. With a 12-year half-life, \( A(t) = A_0\left( \dfrac{1}{2} \right)^{t/12} \). Set up the equation. Ten percent remaining means \( \dfrac{A}{A_0} = 0.10 \): \( \left( \dfrac{1}{2} \right)^{t/12} = 0.10 \). Take natural logarithms. \( \dfrac{t}{12}\ln(0.5) = \ln(0.10) \). \( \ln(0.5) \approx -0.6931 \) and \( \ln(0.10) \approx -2.3026 \). \( \dfrac{t}{12} = \dfrac{-2.3026}{-0.6931} \approx 3.3219 \). Both logarithms are negative and the quotient is positive, which it must be since the time is positive. A negative answer here would signal a setup error. \( t \approx 12 \times 3.3219 \approx 39.86 \) years. Check against the half-life directly. After 36 years, three half-lives have passed, leaving \( \left( \dfrac{1}{2} \right)^3 = 12.5 \) percent. After 48 years, four half-lives, leaving \( 6.25 \) percent. Ten percent lies between 12.5 and 6.25, so the answer must lie between 36 and 48 years. The computed 39.86 does. Consistent. Verify by substitution. \( \left( \dfrac{1}{2} \right)^{39.86/12} = \left( \dfrac{1}{2} \right)^{3.3217} = e^{3.3217 \times (-0.6931)} = e^{-2.3023} \approx 0.1000 \). Correct. Note what the exponent 3.3219 means. It is the number of half-lives required, and it equals \( \log_2 10 \), which appeared in lesson 7.4. Reducing to a tenth takes about 3.32 halvings, whatever the substance, because the ratio is what determines it. A useful consequence. Since ten percent takes 3.32 half-lives, one percent takes twice that, about 6.64 half-lives or 80 years here. Each further factor of ten costs the same 3.32 half-lives, which is the logarithmic scale of lesson 7.7 showing up in a decay problem. About 39.9 years

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Check every candidate.

1. Solve \( \log_3 x = 4 \).
   Show the full solution

   \( x = 3^4 \). \( x = 81 \)

2. Solve \( \log x = 2 \).
   Show the full solution

   Base 10. \( x = 100 \)

3. Solve \( \ln x = 0 \).
   Show the full solution

   \( x = e^0 \). \( x = 1 \)

4. Solve \( \log_2(x + 1) = 3 \).
   Show the full solution

   \( x + 1 = 8 \). \( x = 7 \)

5. What must be true of every logarithm's argument?
   Show the full solution

   It must be positive

6. Solve \( \log_5 x + \log_5 4 = 2 \).
   Show the full solution

   Domain: \( x \gt 0 \). Condense: \( \log_5(4x) = 2 \). Convert: \( 4x = 25 \), so \( x = 6.25 \). Domain check: positive. Genuine. Verify: \( \log_5 6.25 + \log_5 4 = \log_5 25 = 2 \). Correct. \( x = 6.25 \)

7. Solve \( \log(x + 3) = \log(2x - 1) \).
   Show the full solution

   Domain: \( x + 3 \gt 0 \) gives \( x \gt -3 \); \( 2x - 1 \gt 0 \) gives \( x \gt 0.5 \). Binding: \( x \gt 0.5 \). Both sides are single logarithms with the same base, so equate the arguments: \( x + 3 = 2x - 1 \), giving \( x = 4 \). Domain check: \( 4 \gt 0.5 \). Genuine. Verify: \( \log 7 = \log 7 \). Correct. \( x = 4 \)

8. Solve \( \log_2 x + \log_2(x - 2) = 3 \).
   Show the full solution

   Domain: \( x \gt 0 \) and \( x \gt 2 \). Binding: \( x \gt 2 \). Condense: \( \log_2\big(x(x-2)\big) = 3 \). Convert: \( x^2 - 2x = 8 \), so \( x^2 - 2x - 8 = 0 \). Factor: \( (x - 4)(x + 2) = 0 \), giving \( x = 4 \) or \( x = -2 \). \( x = 4 \) satisfies \( x \gt 2 \). Genuine. \( x = -2 \) violates it. Extraneous. Verify: \( \log_2 4 + \log_2 2 = 2 + 1 = 3 \). Correct. \( x = 4 \) only

9. Explain why an extraneous root here is the same phenomenon as in lessons 4.5 and 5.5.
   Show the full solution

   In all three the solving step produces an equation that is implied by the original but does not imply it back, so the new solution set can be strictly larger. Lesson 4.5. Multiplying by a variable denominator is valid only when that denominator is nonzero. At the excluded values it multiplies by zero, which makes any equation true. Lesson 5.5. Squaring is not one-to-one, so it collapses \( B \) and \( -B \) together and accepts solutions where the signs disagreed. This lesson. Condensing logarithms uses properties that hold only for positive arguments. Applying them where an argument is negative produces a valid-looking equation with a larger solution set. The common structure. Each step is a one-way implication. Original implies transformed; transformed does not imply original. So the candidates found form a superset of the true solutions, and checking is a filtering step. The common remedy. Record the conditions from the original before transforming, and test every candidate against the original rather than against any later line. Why it is reassuring. Since no solution is ever lost, the method is safe. The only risk is admitting too many, and that is entirely fixable by checking. All three apply a step that is implied by the original but not equivalent to it, so the solution set can only grow

10. Solve \( \log(x - 1) + \log(x + 2) = 1 \), checking every candidate carefully.
    Show the full solution

    Note the domain from the original. \( x - 1 \gt 0 \) gives \( x \gt 1 \). \( x + 2 \gt 0 \) gives \( x \gt -2 \). The binding condition is \( x \gt 1 \). Condense. By the product property, \( \log\big((x-1)(x+2)\big) = 1 \). Convert to exponential form. The base is 10, so \( (x - 1)(x + 2) = 10 \). Expand and solve. \( x^2 + 2x - x - 2 = 10 \), so \( x^2 + x - 12 = 0 \). Factor: two numbers multiplying to \( -12 \) and adding to 1 are 4 and \( -3 \): \( (x + 4)(x - 3) = 0 \). Candidates: \( x = -4 \) and \( x = 3 \). Apply the domain check. \( x = 3 \) satisfies \( x \gt 1 \). Genuine. \( x = -4 \) violates it. Substituting confirms: \( \log(-5) \) and \( \log(-2) \) are both undefined. Extraneous. Why the condensed form would have passed it. At \( x = -4 \) the condensed argument is \( (-5)(-2) = 10 \), which is positive, so \( \log 10 = 1 \) holds. The condensed equation is genuinely satisfied by \( -4 \); the original is not. That is precisely the hazard the domain note guards against. Verify the survivor in the original. \( \log(3 - 1) + \log(3 + 2) = \log 2 + \log 5 = \log 10 = 1 \). Correct. Numerically. \( 0.3010 + 0.6990 = 1.0000 \). Correct. A note on how many solutions to expect. The condensed equation was quadratic, so at most two candidates. One survived. There is no general rule that a logarithmic equation has one solution, so both candidates always have to be examined rather than assuming the negative one fails. \( x = 3 \) only

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A one-unit increase on a logarithmic scale means what factor?
   Show the full solution

   10

2. Find \( [\text{H}^+] \) for a solution of pH 5.
   Show the full solution

   \( 10^{-5} \). \( 10^{-5} \) moles per liter

3. Which is more acidic, pH 2 or pH 6?
   Show the full solution

   Lower pH means more acidic. pH 2

4. How many times greater is the amplitude of a magnitude 6 earthquake than magnitude 4?
   Show the full solution

   \( 10^2 \). 100 times

5. An increase of 10 dB corresponds to what factor in intensity?
   Show the full solution

   10

6. A solution has \( [\text{H}^+] = 4 \times 10^{-5} \). Find its pH.
   Show the full solution

   \( \text{pH} = -\log(4 \times 10^{-5}) \). Using the product property: \( -\big(\log 4 + \log 10^{-5}\big) = -(0.6021 - 5) = 4.3979 \). About pH 4.40. Check: \( 10^{-4.40} \approx 3.98 \times 10^{-5} \). Correct to rounding. Sanity check: the concentration is between \( 10^{-5} \) and \( 10^{-4} \), so the pH must be between 4 and 5. It is. About 4.40

7. A sound is 85 dB and another is 55 dB. Find the ratio of their intensities.
   Show the full solution

   Difference: \( 85 - 55 = 30 \) dB. Ratio: \( 10^{30/10} = 10^3 = 1000 \). The 85 dB sound has a thousand times the intensity. For reference, 85 dB is heavy traffic and 55 dB is a quiet office, and prolonged exposure above 85 dB is the usual threshold for hearing protection. 1000 times

8. An earthquake of magnitude 6.5 is compared with one of magnitude 5.0. Find the amplitude ratio.
   Show the full solution

   Difference: \( 6.5 - 5.0 = 1.5 \). Ratio: \( 10^{1.5} = 10 \times 10^{0.5} = 10\sqrt{10} \approx 31.6 \). The magnitude 6.5 quake has about 31.6 times the wave amplitude. A non-integer difference still works, and \( 10^{0.5} = \sqrt{10} \) is worth recognizing. Check: \( \log(31.6) \approx 1.4997 \), close to 1.5. Correct. About 31.6 times

9. Explain why a magnitude 8 earthquake is far more than twice as severe as a magnitude 4.
   Show the full solution

   The magnitudes are logarithms, so the ratio of the underlying quantities is determined by the difference of the magnitudes, not their quotient. The amplitude comparison. The difference is \( 8 - 4 = 4 \), so the amplitude ratio is \( 10^4 = 10{,}000 \). The ground moves ten thousand times as far, not twice as far. The energy comparison. Energy scales as roughly \( 10^{1.5} \) per magnitude unit, so the ratio is \( 10^{1.5 \times 4} = 10^6 \), a million times the energy released. Why "twice" is the wrong instinct. Dividing 8 by 4 treats the scale as linear, which is exactly what it is not. On a logarithmic scale the arithmetic operation that means "how many times bigger" is subtraction followed by exponentiation, never division. A check that makes it concrete. Magnitude 4 quakes happen thousands of times a year and are usually felt but harmless. Magnitude 8 quakes occur roughly once a year worldwide and are catastrophic. If the relationship were a factor of two, that difference in consequence would be inexplicable. The general rule. To compare two readings on any logarithmic scale, subtract and then raise the base to the result. Subtracting gives 4, so the amplitude ratio is \( 10^4 \) and the energy ratio about \( 10^6 \)

10. Rainwater has pH 5.6 because dissolved carbon dioxide makes it slightly acidic. Acid rain in an industrial region measures pH 4.2. Find how many times more acidic it is, and comment on whether the scale makes the difference look larger or smaller than it is.
    Show the full solution

    Convert both to concentrations. Normal rain: \( [\text{H}^+] = 10^{-5.6} \). Acid rain: \( [\text{H}^+] = 10^{-4.2} \). Find the ratio. \( \dfrac{10^{-4.2}}{10^{-5.6}} = 10^{-4.2 - (-5.6)} = 10^{1.4} \). \( 10^{1.4} = 10 \times 10^{0.4} \approx 10 \times 2.512 \approx 25.1 \). The acid rain is about 25 times more acidic. Verify with actual concentrations. \( 10^{-5.6} \approx 2.51 \times 10^{-6} \) and \( 10^{-4.2} \approx 6.31 \times 10^{-5} \). Ratio: \( \dfrac{6.31 \times 10^{-5}}{2.51 \times 10^{-6}} \approx 25.1 \). Agrees. Does the scale make it look larger or smaller? Much smaller. On the pH scale the difference reads as 1.4, a change that sounds minor and would be easy to dismiss. The underlying change is a factor of 25, which is substantial. Why that matters here specifically. A 25-fold increase in hydrogen ion concentration is enough to damage forests, acidify lakes past the tolerance of fish eggs, and dissolve limestone and marble. Reporting only the pH difference of 1.4 understates the effect to anyone who reads the scale linearly. The honest way to report it. Give both: the pH values, because that is the standard measurement, and the concentration ratio, because that is what the chemistry responds to. A report that gives only one of the two is incomplete. A note on the reference point. Pure water is pH 7, but unpolluted rain is naturally about 5.6 because atmospheric carbon dioxide dissolves into it forming carbonic acid. So acid rain is defined against 5.6, not 7, and using 7 as the baseline would overstate the human contribution. About 25 times more acidic; the scale makes a substantial difference look small

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. For \( a_n = 2n + 1 \), find \( a_5 \).
   Show the full solution

   \( 2(5) + 1 \). 11

2. For \( a_n = n^2 \), list the first four terms.
   Show the full solution

   1, 4, 9, 16

3. For \( a_1 = 3 \), \( a_n = a_{n-1} + 5 \), find \( a_3 \).
   Show the full solution

   \( a_2 = 8 \), \( a_3 = 13 \). 13

4. What must every recursive rule include besides the recursion?
   Show the full solution

   A starting value

5. What is the domain of a sequence?
   Show the full solution

   The positive integers

6. Write an explicit rule for 6, 10, 14, 18, and find \( a_{30} \).
   Show the full solution

   The terms increase by 4, so the rule has the form \( 4n + c \). At \( n = 1 \): \( 4 + c = 6 \), so \( c = 2 \). \( a_n = 4n + 2 \). Check: \( n = 3 \) gives 14 ✓, \( n = 4 \) gives 18 ✓. \( a_{30} = 4(30) + 2 = 122 \). \( a_n = 4n + 2 \), \( a_{30} = 122 \)

7. Write a recursive rule for 2, 6, 18, 54.
   Show the full solution

   Each term is 3 times the previous: \( 6 = 3(2) \), \( 18 = 3(6) \), \( 54 = 3(18) \). \( a_1 = 2 \), \( a_n = 3a_{n-1} \). Multiplication, not addition, which is what makes this geometric rather than arithmetic. \( a_1 = 2 \), \( a_n = 3a_{n-1} \)

8. For \( a_1 = 100 \) and \( a_n = 0.5a_{n-1} + 10 \), find the first five terms and describe what happens.
   Show the full solution

   \( a_1 = 100 \). \( a_2 = 0.5(100) + 10 = 60 \). \( a_3 = 0.5(60) + 10 = 40 \). \( a_4 = 0.5(40) + 10 = 30 \). \( a_5 = 0.5(30) + 10 = 25 \). The terms are falling but the gaps are shrinking: 40, 20, 10, 5. They are approaching a limit. Find it. A steady value \( L \) satisfies \( L = 0.5L + 10 \), so \( 0.5L = 10 \) and \( L = 20 \). Check: \( 0.5(20) + 10 = 20 \). Once the sequence reaches 20 it stays there. The terms approach 20 from above without reaching it. This is the drug dosage pattern of lesson 6.7 appearing as a recursion. 100, 60, 40, 30, 25, approaching 20

9. Explain why a recursive rule needs a starting value but an explicit rule does not.
   Show the full solution

   An explicit rule computes each term from its position alone, while a recursive rule computes each term from the one before, so it needs somewhere to begin. The explicit case. In \( a_n = 3n + 2 \), asking for \( a_7 \) needs only the number 7. Nothing earlier in the sequence is consulted, so no anchor is required. The recursive case. In \( a_n = a_{n-1} + 3 \), computing \( a_7 \) requires \( a_6 \), which requires \( a_5 \), and so on back to \( a_1 \). If \( a_1 \) is not given, the chain has no bottom and nothing can be computed. Concretely. The rule \( a_n = a_{n-1} + 3 \) is satisfied by 5, 8, 11 and by 100, 103, 106 and by infinitely many other sequences. The starting value is what selects one of them. The parallel elsewhere. This is the same structure as an antiderivative needing an initial condition, or a differential equation needing a boundary value. A rule about change determines the shape but not the position. How many starting values. A recursion referring only to \( a_{n-1} \) needs one starting value. Fibonacci refers to \( a_{n-1} \) and \( a_{n-2} \), so it needs two. Recursion computes from the previous term, so without an anchor the chain never terminates

10. A sequence begins 1, 2, 4, 8. Give two different rules that produce these four terms and disagree at the fifth.
    Show the full solution

    The obvious rule. Doubling: \( a_n = 2^{n-1} \). Terms: 1, 2, 4, 8, and the fifth is \( 2^4 = 16 \). A second rule that also fits. The number of regions a circle is cut into by chords joining \( n \) points on its edge, in general position, is \[ a_n = \frac{n^4 - 6n^3 + 23n^2 - 18n + 24}{24} \] Check \( n = 1 \): \( \dfrac{1 - 6 + 23 - 18 + 24}{24} = \dfrac{24}{24} = 1 \) ✓ Check \( n = 2 \): \( \dfrac{16 - 48 + 92 - 36 + 24}{24} = \dfrac{48}{24} = 2 \) ✓ Check \( n = 3 \): \( \dfrac{81 - 162 + 207 - 54 + 24}{24} = \dfrac{96}{24} = 4 \) ✓ Check \( n = 4 \): \( \dfrac{256 - 384 + 368 - 72 + 24}{24} = \dfrac{192}{24} = 8 \) ✓ Where they disagree. At \( n = 5 \): \( \dfrac{625 - 750 + 575 - 90 + 24}{24} = \dfrac{384}{24} = 16 \). Still 16. They agree at the fifth term too. At \( n = 6 \): \( \dfrac{1296 - 1296 + 828 - 108 + 24}{24} = \dfrac{744}{24} = 31 \), while doubling gives 32. They disagree at the sixth. A simpler pair that disagrees at the fifth. Take \( b_n = 2^{n-1} \) and \( c_n = 2^{n-1} + (n-1)(n-2)(n-3)(n-4) \). The added product is zero at \( n = 1, 2, 3, 4 \), so the first four terms match. At \( n = 5 \) it contributes \( 4 \cdot 3 \cdot 2 \cdot 1 = 24 \), giving 40 instead of 16. What this shows. Four terms never determine a sequence. Adding any expression that vanishes at those four positions produces another rule fitting the same data. There are infinitely many such rules. Why the convention is still reasonable. When a problem asks for "the" pattern, it wants the simplest rule that fits, and in practice that is nearly always intended and nearly always unique in its simplicity. But "the next term must be 16" is a statement about convention, not about logic. Why it matters practically. In any setting where the sequence comes from data rather than from a stated rule, extrapolating past the observed terms is a guess. The mathematics does not certify it. \( a_n = 2^{n-1} \) and \( a_n = 2^{n-1} + (n-1)(n-2)(n-3)(n-4) \), which give 16 and 40

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the common difference of 3, 9, 15, 21.
   Show the full solution

   6

2. Find \( a_{10} \) for \( a_1 = 5 \), \( d = 3 \).
   Show the full solution

   \( 5 + 9(3) \). 32

3. Write the explicit rule for \( a_1 = 2 \), \( d = 7 \).
   Show the full solution

   \( 2 + 7(n-1) = 7n - 5 \). \( a_n = 7n - 5 \)

4. Is 4, 8, 16, 32 arithmetic?
   Show the full solution

   Differences 4, 8, 16 are not constant. No

5. Find the common difference of 20, 17, 14, 11.
   Show the full solution

   \( -3 \)

6. An arithmetic sequence has \( a_3 = 14 \) and \( a_7 = 30 \). Find \( a_1 \) and \( d \).
   Show the full solution

   From position 3 to position 7 is 4 steps, and the value rose by \( 30 - 14 = 16 \). So \( 4d = 16 \) and \( d = 4 \). Step back from \( a_3 \) by two: \( a_1 = 14 - 2(4) = 6 \). Check: 6, 10, 14, 18, 22, 26, 30. The third is 14 ✓ and the seventh is 30 ✓. \( a_1 = 6 \), \( d = 4 \)

7. Determine whether 87 appears in 5, 12, 19, 26.
   Show the full solution

   \( d = 7 \), so \( a_n = 5 + 7(n-1) = 7n - 2 \). Set \( 7n - 2 = 87 \): \( 7n = 89 \), \( n = \dfrac{89}{7} \approx 12.71 \). Not an integer, so 87 is not a term. Confirm: \( a_{12} = 82 \) and \( a_{13} = 89 \). The sequence steps over 87. No

8. A theater has 22 seats in the first row and 3 more in each successive row. Find the number of seats in row 15 and which row has 76 seats.
   Show the full solution

   Arithmetic with \( a_1 = 22 \) and \( d = 3 \). \( a_n = 22 + 3(n-1) = 3n + 19 \). Row 15: \( a_{15} = 3(15) + 19 = 64 \) seats. Row with 76 seats: \( 3n + 19 = 76 \), so \( 3n = 57 \) and \( n = 19 \). Check: \( 3(19) + 19 = 76 \) ✓. 64 seats in row 15; row 19 has 76 seats

9. Explain why an arithmetic sequence is a linear function and what the restriction changes.
   Show the full solution

   Both are defined by a constant rate of change, and the restriction is only on which inputs are allowed. The shared structure. A linear function adds \( m \) to the output for every unit added to the input. An arithmetic sequence adds \( d \) for every step in position. Those are the same statement, with \( d \) playing the role of \( m \). The algebra confirms it. Expanding \( a_n = a_1 + (n-1)d \) gives \( a_n = dn + (a_1 - d) \), which is \( y = mx + b \) with \( m = d \) and \( b = a_1 - d \). What the restriction changes. The domain. A linear function accepts every real input; a sequence accepts only positive integers. So the graph is isolated dots rather than a connected line. The practical consequence. A value can lie on the line without being a term, which is exactly what happens when solving for a position gives a fraction. The linear equation has a solution; the sequence question does not. What carries over. Slope, intercept, and solving for an input all work the same way. Only the final check on whether the answer is a legal position is new. Both have a constant rate of change; restricting the domain to positive integers turns the line into isolated points

10. The sum of the first three terms of an arithmetic sequence is 24 and the sum of the next three is 69. Find the first term and the common difference.
    Show the full solution

    Write the first six terms in terms of \( a_1 \) and \( d \). \( a_1 \), \( a_1 + d \), \( a_1 + 2d \), \( a_1 + 3d \), \( a_1 + 4d \), \( a_1 + 5d \). Translate the first condition. \( a_1 + (a_1 + d) + (a_1 + 2d) = 3a_1 + 3d = 24 \). Dividing by 3: \( a_1 + d = 8 \). Translate the second condition. \( (a_1 + 3d) + (a_1 + 4d) + (a_1 + 5d) = 3a_1 + 12d = 69 \). Dividing by 3: \( a_1 + 4d = 23 \). Solve the system. Subtracting the first from the second: \( 3d = 15 \), so \( d = 5 \). Then \( a_1 + 5 = 8 \), giving \( a_1 = 3 \). Check both conditions. The sequence is 3, 8, 13, 18, 23, 28. First three: \( 3 + 8 + 13 = 24 \) ✓ Next three: \( 18 + 23 + 28 = 69 \) ✓ A shortcut worth noticing. Dividing each sum by 3 gave the middle term of that group: \( a_2 = 8 \) and \( a_5 = 23 \). In an arithmetic sequence the mean of any odd number of consecutive terms equals the middle term, because the deviations above and below cancel in pairs. Using that directly. From \( a_2 = 8 \) and \( a_5 = 23 \), three steps apart, \( 3d = 15 \) so \( d = 5 \), and \( a_1 = 8 - 5 = 3 \). Same answer in two lines. \( a_1 = 3 \), \( d = 5 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( 1 + 2 + \cdots + 10 \).
   Show the full solution

   \( \dfrac{10}{2}(1 + 10) \). 55

2. Find \( S_{10} \) for \( a_1 = 3 \), \( a_{10} = 30 \).
   Show the full solution

   \( 5(33) \). 165

3. How many terms are in \( \displaystyle\sum_{k=1}^{15} \)?
   Show the full solution

   15

4. What is the difference between a sequence and a series?
   Show the full solution

   A sequence is a list; a series is the sum of its terms

5. Evaluate \( \displaystyle\sum_{k=1}^{4} 2k \).
   Show the full solution

   \( 2 + 4 + 6 + 8 \). 20

6. Find the sum of the first 25 terms of 4, 9, 14, 19.
   Show the full solution

   \( a_1 = 4 \), \( d = 5 \). \( a_{25} = 4 + 24(5) = 124 \). \( S_{25} = \dfrac{25}{2}(4 + 124) = \dfrac{25}{2}(128) = 25(64) = 1600 \). Check with the other form: \( \dfrac{25}{2}(8 + 120) = \dfrac{25}{2}(128) = 1600 \). Agrees. 1600

7. Evaluate \( \displaystyle\sum_{k=5}^{12} (3k - 1) \).
   Show the full solution

   Count the terms. From 5 to 12 inclusive is \( 12 - 5 + 1 = 8 \) terms. Find the first and last. At \( k = 5 \): \( 3(5) - 1 = 14 \). At \( k = 12 \): \( 3(12) - 1 = 35 \). Apply the formula. \( S = \dfrac{8}{2}(14 + 35) = 4(49) = 196 \). Check by listing: 14, 17, 20, 23, 26, 29, 32, 35. Sum in pairs from the outside: \( 14 + 35 = 49 \), \( 17 + 32 = 49 \), \( 20 + 29 = 49 \), \( 23 + 26 = 49 \). Four pairs of 49 give 196 ✓ The \( +1 \) in the count is essential; using 7 terms would give 171.5, which is not even a whole number and so is visibly wrong. 196

8. A worker is paid $500 in week 1 with a $25 raise each week. Find the total pay over one year of 52 weeks.
   Show the full solution

   Arithmetic with \( a_1 = 500 \), \( d = 25 \), \( n = 52 \). Final week: \( a_{52} = 500 + 51(25) = 500 + 1275 = 1775 \). Total: \( S_{52} = \dfrac{52}{2}(500 + 1775) = 26(2275) = 59{,}150 \). Sanity check: the average weekly pay is \( \dfrac{500 + 1775}{2} = 1137.50 \), and \( 52 \times 1137.50 = 59{,}150 \) ✓ Note the size of the raise's effect. Flat pay at $500 would total $26,000. The raises more than doubled the year's earnings. $59,150

9. Explain why the pairing derivation works and what would break it for a geometric sequence.
   Show the full solution

   Pairing works because in an arithmetic sequence moving inward from both ends changes the two terms by equal and opposite amounts. The mechanism. Stepping forward one place from the front adds \( d \). Stepping back one place from the end subtracts \( d \). So the pair's sum is unchanged, and every pair equals \( a_1 + a_n \). Why that gives the formula. Writing the sum forward and backward and adding produces \( n \) identical pairs, so \( 2S_n = n(a_1 + a_n) \). Why it fails for geometric. There, stepping forward multiplies by \( r \) and stepping back divides by \( r \). Those do not cancel in a sum. Take 1, 2, 4, 8: the outer pair gives \( 1 + 8 = 9 \) and the inner pair gives \( 2 + 4 = 6 \). Not equal, so the argument collapses at its first step. What replaces it. Lesson 8.5 uses a different trick entirely: multiplying the whole sum by \( r \) and subtracting, which exploits the multiplicative structure instead of fighting it. The general lesson. Each kind of series needs a derivation matched to how its terms are built. There is no single technique that handles both. Equal and opposite steps cancel in an arithmetic pair; in a geometric sequence the steps multiply and divide, which does not cancel in a sum

10. The sum of the first \( n \) terms of an arithmetic sequence with \( a_1 = 6 \) and \( d = 4 \) is 510. Find \( n \).
    Show the full solution

    Use the form that does not need \( a_n \). \( S_n = \dfrac{n}{2}\big(2a_1 + (n-1)d\big) = \dfrac{n}{2}\big(12 + 4(n-1)\big) \). Simplify inside. \( 12 + 4n - 4 = 4n + 8 \). So \( S_n = \dfrac{n(4n + 8)}{2} = \dfrac{4n^2 + 8n}{2} = 2n^2 + 4n \). Set equal to 510 and solve. \( 2n^2 + 4n = 510 \). \( 2n^2 + 4n - 510 = 0 \). Divide by 2: \( n^2 + 2n - 255 = 0 \). Factor. Two numbers multiplying to \( -255 \) and adding to 2. Since \( 255 = 15 \times 17 \), try 17 and \( -15 \): product \( -255 \) ✓, sum 2 ✓. \( (n + 17)(n - 15) = 0 \), so \( n = -17 \) or \( n = 15 \). Reject the negative. A count of terms must be a positive integer, so \( n = 15 \). This rejection is not an extraneous root in the technical sense. The algebra is exact; \( -17 \) is a genuine solution of the quadratic. It is rejected because the context restricts the domain, which is a different kind of check from the one in lesson 7.6. Verify. \( a_{15} = 6 + 14(4) = 62 \). \( S_{15} = \dfrac{15}{2}(6 + 62) = \dfrac{15}{2}(68) = 15(34) = 510 \) ✓ Note the quadratic structure. The sum \( S_n = 2n^2 + 4n \) is quadratic in \( n \), which is always true of an arithmetic series: adding a linear amount repeatedly accumulates quadratically. That is why questions about a total can have two algebraic answers when questions about a term cannot. \( n = 15 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the common ratio of 2, 10, 50, 250.
   Show the full solution

   5

2. Find \( a_5 \) for \( a_1 = 4 \), \( r = 3 \).
   Show the full solution

   \( 4 \cdot 3^4 = 4(81) \). 324

3. Write the explicit rule for \( a_1 = 5 \), \( r = 2 \).
   Show the full solution

   \( a_n = 5 \cdot 2^{n-1} \)

4. Is 3, 7, 11, 15 geometric?
   Show the full solution

   Ratios are not constant; the differences are. No, it is arithmetic

5. Find the common ratio of 80, 40, 20, 10.
   Show the full solution

   \( \frac{1}{2} \)

6. Find \( a_8 \) for the sequence 2, \( -6 \), 18, \( -54 \).
   Show the full solution

   \( r = \dfrac{-6}{2} = -3 \). Confirm: \( \dfrac{18}{-6} = -3 \) ✓ \( a_8 = 2(-3)^7 = 2(-2187) = -4374 \). The sign alternates because \( r \) is negative, and odd powers keep the minus sign. Even positions are negative here and odd positions positive, since \( n - 1 \) is odd when \( n \) is even. \( -4374 \)

7. A geometric sequence has \( a_2 = 12 \) and \( a_5 = 96 \). Find \( a_1 \) and \( r \).
   Show the full solution

   From position 2 to position 5 is 3 steps, so \( a_5 = a_2 \cdot r^3 \). \( 96 = 12r^3 \), so \( r^3 = 8 \) and \( r = 2 \). Step back one: \( a_1 = \dfrac{12}{2} = 6 \). Check: 6, 12, 24, 48, 96. Second is 12 ✓ and fifth is 96 ✓ Note the contrast with the arithmetic version. There, three steps meant \( 3d \); here it means \( r^3 \). Steps add in one case and multiply in the other. \( a_1 = 6 \), \( r = 2 \)

8. A ball dropped from 10 feet rebounds to 60 percent of its previous height each bounce. Find the height after the fifth bounce.
   Show the full solution

   Geometric with \( a_1 = 10 \) (the drop height) and \( r = 0.6 \). After the fifth bounce the ball has been multiplied by 0.6 five times: \( 10(0.6)^5 \). \( 0.6^2 = 0.36 \), \( 0.6^3 = 0.216 \), \( 0.6^4 = 0.1296 \), \( 0.6^5 = 0.07776 \). Height \( = 10(0.07776) = 0.7776 \) feet, about 9.3 inches. Watch the indexing. Calling the drop height \( a_1 \) makes the height after the first bounce \( a_2 \), so "after the fifth bounce" is \( a_6 = 10(0.6)^5 \). Stating which term is which before computing avoids an off-by-one error here. About 0.778 feet

9. Explain why a geometric sequence is an exponential function and what the restriction changes.
   Show the full solution

   Both are defined by a constant multiplier per step, and again only the domain differs. The shared structure. An exponential function multiplies the output by \( b \) for every unit added to the input. A geometric sequence multiplies by \( r \) for every step in position. The same statement, with \( r \) as the base. The algebra. \( a_n = a_1 r^{n-1} = \dfrac{a_1}{r} \cdot r^n \), which is \( f(x) = ab^x \) with \( b = r \) and \( a = \dfrac{a_1}{r} \). What the restriction changes. Only positive integers are legal inputs, so the graph is isolated points on the exponential curve rather than the curve itself. What carries over from unit 6. Everything about growth and decay: \( r \gt 1 \) grows, \( 0 \lt r \lt 1 \) decays, doubling and half-life arguments apply unchanged, and solving for a position uses the logarithms of lesson 7.5. One thing that does not carry over. A negative common ratio makes a perfectly good sequence with alternating signs, while \( b^x \) with a negative base is not a real function for most inputs. So the sequence version is slightly more general at that point. Both multiply by a constant per step; restricting the domain to positive integers turns the curve into isolated points

10. A culture starts with 500 bacteria and triples every 4 hours. Write the count as a geometric sequence indexed by 4-hour periods, find the count after 24 hours, and find when it first exceeds one million.
    Show the full solution

    Set up the indexing. Let \( a_1 = 500 \) be the count at time zero, so \( a_n \) is the count after \( n - 1 \) periods of 4 hours. \( a_n = 500 \cdot 3^{n-1} \). Find the count after 24 hours. That is \( \dfrac{24}{4} = 6 \) periods, so \( n - 1 = 6 \) and \( n = 7 \). \( a_7 = 500 \cdot 3^6 = 500(729) = 364{,}500 \). Check by stepping. 500, 1500, 4500, 13500, 40500, 121500, 364500. Seven terms, the last at 24 hours ✓ Find when it first exceeds one million. \( 500 \cdot 3^{n-1} \gt 1{,}000{,}000 \). Isolate: \( 3^{n-1} \gt 2000 \). Take natural logarithms: \( (n-1)\ln 3 \gt \ln 2000 \). \( n - 1 \gt \dfrac{7.6009}{1.0986} \approx 6.919 \). So \( n \gt 7.919 \), and since \( n \) must be an integer, \( n = 8 \). Check both sides of the threshold. \( a_7 = 364{,}500 \), which is below one million. \( a_8 = 500 \cdot 3^7 = 500(2187) = 1{,}093{,}500 \), which is above ✓ Convert to time. \( n = 8 \) means 7 periods, or \( 7 \times 4 = 28 \) hours. Why rounding up is right here. The inequality asked for the first term exceeding the threshold, so a fractional answer must be rounded up to the next integer position. Rounding 7.919 down to 7 would give a term still below one million. A note on the continuous version. The population is actually growing continuously, not jumping every 4 hours, so the true crossing happens partway through the eighth period. The sequence model reports the first observation past the threshold, which is the right answer to the question as asked but not the same as the exact crossing time. 364,500 after 24 hours; it first exceeds one million at 28 hours

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( 1 + 2 + 4 + 8 \) using the formula.
   Show the full solution

   \( \dfrac{2^4 - 1}{2 - 1} = 15 \). 15

2. Find \( S_5 \) for \( a_1 = 2 \), \( r = 3 \).
   Show the full solution

   \( 2 \cdot \dfrac{243 - 1}{2} = 242 \). 242

3. Which value of \( r \) must be excluded from the formula?
   Show the full solution

   \( r = 1 \)

4. Does the sum formula use \( r^n \) or \( r^{n-1} \)?
   Show the full solution

   \( r^n \)

5. Find \( S_4 \) for \( a_1 = 5 \), \( r = 2 \).
   Show the full solution

   \( 5(15) \). Check: \( 5 + 10 + 20 + 40 = 75 \) ✓ 75

6. Find the sum of the first 6 terms of 4, 12, 36.
   Show the full solution

   \( a_1 = 4 \), \( r = 3 \), \( n = 6 \). \( S_6 = 4 \cdot \dfrac{3^6 - 1}{3 - 1} = 4 \cdot \dfrac{729 - 1}{2} = 4(364) = 1456 \). Check by listing: 4, 12, 36, 108, 324, 972. Sum: \( 4 + 12 + 36 + 108 + 324 + 972 = 1456 \) ✓ 1456

7. Find the sum of the first 10 terms of 64, 32, 16.
   Show the full solution

   \( a_1 = 64 \), \( r = 0.5 \), \( n = 10 \). \( 0.5^{10} = \dfrac{1}{1024} \approx 0.0009766 \). \( S_{10} = 64 \cdot \dfrac{1 - 0.0009766}{0.5} = 64(1.9980469) = 127.875 \). Exactly: \( 64 \cdot \dfrac{1 - \frac{1}{1024}}{\frac{1}{2}} = 128 \cdot \dfrac{1023}{1024} = \dfrac{1023}{8} = 127.875 \). The sum is creeping toward 128 and is now within \( \frac{1}{8} \) of it. 127.875

8. A machine worth $40,000 loses 15 percent of its value each year. Find the total depreciation over 5 years.
   Show the full solution

   Find each year's loss. Year 1 loses \( 0.15(40000) = 6000 \), leaving $34,000. Year 2 loses \( 0.15(34000) = 5100 \). Each loss is 0.85 times the previous, so the losses form a geometric sequence with \( a_1 = 6000 \) and \( r = 0.85 \). Sum five terms. \( 0.85^5 = 0.4437053 \). \( S_5 = 6000 \cdot \dfrac{1 - 0.4437053}{0.15} = 6000 \cdot \dfrac{0.5562947}{0.15} = 6000(3.708631) = 22{,}251.79 \). Check a different way. The value after 5 years is \( 40000(0.85)^5 = 40000(0.4437053) = 17{,}748.21 \). Total depreciation \( = 40000 - 17748.21 = 22{,}251.79 \) ✓ The second route is shorter, and it is worth noticing that a question about accumulated loss can often be answered by tracking the remaining value instead. About $22,251.79

9. Explain why the shift-and-subtract trick works and why it does not help an arithmetic series.
   Show the full solution

   Multiplying a geometric sum by \( r \) produces almost the same sum shifted one position, so subtracting cancels everything in the overlap. The mechanism. Each term times \( r \) is the next term. So \( rS_n \) contains terms 2 through \( n+1 \), while \( S_n \) contains terms 1 through \( n \). The overlap is terms 2 through \( n \), which is nearly all of both, and subtracting removes it entirely. What survives. Only \( a_1 \) from the first sum and \( a_1r^n \) from the second, which is why the formula is so short. Why it fails for arithmetic. There, multiplying by anything does not shift the sequence along itself. Take 1, 3, 5, 7. Multiplying by 3 gives 3, 9, 15, 21, which is a different arithmetic sequence, not this one shifted. Nothing cancels on subtraction. What does shift an arithmetic sequence. Adding \( d \) to every term gives 3, 5, 7, 9, which is the original shifted. Subtracting \( S_n \) from that shifted sum gives \( nd \), which recovers the sum formula by a slightly different route than pairing. The general principle. The operation that shifts a sequence along itself is the operation that built it: addition for arithmetic, multiplication for geometric. Each derivation uses its own. Multiplying by \( r \) shifts a geometric sequence one place, so subtraction cancels the overlap; multiplication does not shift an arithmetic sequence

10. A person saves $100 at the end of each month in an account paying 4.8 percent annual interest compounded monthly. Find the balance after 3 years, and find how much of it is interest.
    Show the full solution

    Find the monthly rate and the number of deposits. Monthly rate: \( \dfrac{0.048}{12} = 0.004 \). Number of deposits: \( 3 \times 12 = 36 \). Set up the series. The last deposit earns nothing, the one before earns one month, and the first earns 35 months. So the total is \[ 100 + 100(1.004) + 100(1.004)^2 + \cdots + 100(1.004)^{35} \] a geometric series with \( a_1 = 100 \), \( r = 1.004 \), \( n = 36 \). Compute the power. \( \ln(1.004) \approx 0.0039920 \). \( 36 \times 0.0039920 = 0.1437129 \). \( 1.004^{36} = e^{0.1437129} \approx 1.1545537 \). Apply the formula. \( S_{36} = 100 \cdot \dfrac{1.1545537 - 1}{0.004} = 100 \cdot \dfrac{0.1545537}{0.004} = 100(38.638425) = 3863.84 \). The balance is about $3,863.84. Separate the interest. Total deposited: \( 36 \times 100 = \$3{,}600 \). Interest: \( 3863.84 - 3600 = \$263.84 \). Check for reasonableness. The average deposit sits about 17.5 months. \( \$3{,}600 \) at 4.8 percent for 17.5 months is roughly \( 3600 \times 0.048 \times \dfrac{17.5}{12} \approx \$252 \). Close to $263.84, with the difference coming from compounding. Consistent. Compare with a lump sum. Depositing all $3,600 at the start would give \( 3600(1.004)^{36} = 3600(1.1545537) = \$4{,}156.39 \), nearly $300 more. Money deposited later has less time to earn, which is why the monthly total falls short of the lump sum. The practical reading. Over three years at this rate the interest is about 7.3 percent of what was deposited. That is modest, and the reason to save monthly is that $3,600 at once is usually not available, not that it earns more. $3,863.84, of which $263.84 is interest

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. For what values of \( r \) does an infinite geometric series converge?
   Show the full solution

   \( |r| \lt 1 \)

2. Find the sum of \( 6 + 3 + 1.5 + \cdots \).
   Show the full solution

   \( \dfrac{6}{1 - 0.5} \). 12

3. Does \( 1 + 2 + 4 + \cdots \) converge?
   Show the full solution

   \( r = 2 \). No

4. Find the sum of \( 100 + 20 + 4 + \cdots \).
   Show the full solution

   \( r = 0.2 \), so \( \dfrac{100}{0.8} \). 125

5. Convert \( 0.\overline{7} \) to a fraction.
   Show the full solution

   \( \dfrac{0.7}{0.9} \). \( \frac{7}{9} \)

6. Find the sum of \( 27 - 9 + 3 - 1 + \cdots \).
   Show the full solution

   \( r = \dfrac{-9}{27} = -\dfrac{1}{3} \), and \( |r| = \dfrac{1}{3} \lt 1 \), so it converges. \( S = \dfrac{27}{1 - \left( -\frac{1}{3} \right)} = \dfrac{27}{\frac{4}{3}} = 27 \cdot \dfrac{3}{4} = \dfrac{81}{4} = 20.25 \). Check with partial sums: 27, 18, 21, 20, 20.333, 20.222. Closing in on 20.25 from both sides ✓ A negative ratio makes the partial sums alternate around the limit rather than approaching from one side. 20.25

7. Convert \( 0.2\overline{45} \) to a fraction.
   Show the full solution

   Split off the non-repeating part. \( 0.2\overline{45} = 0.2 + 0.0454545\ldots \) The repeating part is a geometric series with \( a_1 = 0.045 \) and \( r = 0.01 \). \( S = \dfrac{0.045}{0.99} = \dfrac{45}{990} = \dfrac{1}{22} \). Add the parts. \( \dfrac{1}{5} + \dfrac{1}{22} = \dfrac{22}{110} + \dfrac{5}{110} = \dfrac{27}{110} \). Check: \( 27 \div 110 = 0.24545\ldots \) ✓ The non-repeating digits must be separated first, because they are not part of the geometric pattern. \( \frac{27}{110} \)

8. A ball dropped from 12 feet rebounds to 40 percent of its height each bounce. Find the total vertical distance it travels before coming to rest.
   Show the full solution

   Account for both directions. The ball falls 12 feet, then rises and falls 4.8 feet, then rises and falls 1.92 feet, and so on. Total \( = 12 + 2(4.8 + 1.92 + 0.768 + \cdots) \). Sum the series inside. \( a_1 = 4.8 \), \( r = 0.4 \). \( S = \dfrac{4.8}{1 - 0.4} = \dfrac{4.8}{0.6} = 8 \). Combine. Total \( = 12 + 2(8) = 28 \) feet. The factor of 2 is where this problem is usually lost. Every rebound height is traveled twice, once up and once down, while the initial drop is traveled once. Check the size. The ball falls 12, rises 4.8, falls 4.8, rises 1.92, and so on. Partial totals: 12, 16.8, 21.6, 23.52, 25.44, 26.2. Climbing toward 28 ✓ 28 feet

9. Explain why an infinite sum of positive numbers can be finite.
   Show the full solution

   Because the terms shrink fast enough that the running total is bounded above, and a bounded increasing total has a limit. The mechanism with a specific series. In \( 8 + 4 + 2 + 1 + \cdots \), each partial sum is exactly halfway from the previous one to 16. So the total gets closer to 16 at every step and never passes it. Why "infinitely many" does not force "infinitely large". How many terms there are is not what decides the total. What decides it is how fast they shrink. Adding infinitely many copies of 0.001 gives an unbounded total; adding infinitely many halvings does not. The condition made precise. For a geometric series, shrinking fast enough means \( |r| \lt 1 \). The remaining amount after \( n \) terms is \( a_1r^n \) divided by \( 1 - r \), and that goes to zero. A counterexample worth knowing about. Shrinking terms alone is not enough. The series \( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots \) has terms going to zero and yet grows without bound. It just does so extremely slowly, needing more than \( 10^{43} \) terms to pass 100. Geometric shrinkage is fast enough; that one is not. Calculus proves which is which. The terms shrink fast enough that the running total stays below a fixed bound

10. A drug with a 6-hour half-life is taken as a 200 mg dose every 6 hours. Find the long-run maximum and minimum amounts in the body, using an infinite series.
    Show the full solution

    Set up what remains from each past dose. With a 6-hour half-life and 6-hour intervals, each dose has exactly half left when the next arrives. Find the maximum, just after a dose. In the long run the body holds the new 200 mg, plus half of the previous dose, plus a quarter of the one before, and so on: \[ 200 + 100 + 50 + 25 + \cdots \] This is geometric with \( a_1 = 200 \) and \( r = 0.5 \), and \( |r| \lt 1 \), so it converges. \( S = \dfrac{200}{1 - 0.5} = \dfrac{200}{0.5} = 400 \) mg. Find the minimum, just before the next dose. Six hours later everything has halved: \( \dfrac{400}{2} = 200 \) mg. Check the steady state directly. If the peak is \( P \), then six hours later the body holds \( \dfrac{P}{2} \), and adding the next dose gives \( \dfrac{P}{2} + 200 \). For a steady state this must equal \( P \): \( \dfrac{P}{2} + 200 = P \), so \( \dfrac{P}{2} = 200 \) and \( P = 400 \) ✓ Two independent routes agreeing is the check worth having, since the series argument and the steady-state equation use entirely different reasoning. Trace the approach. Dose 1: peak 200, trough 100. Dose 2: peak 300, trough 150. Dose 3: peak 350, trough 175. Dose 4: peak 375, trough 187.5. Dose 5: peak 387.5. Climbing toward 400, halving the remaining gap each time ✓ What this means clinically. The body holds twice as much at steady state as a single dose provides, and reaching that level takes four or five doses, about a day here. That delay is why some medications begin with a larger loading dose: to reach the working level immediately instead of after a day. The fluctuation. The amount swings between 400 and 200 mg, a factor of two. Shorter intervals relative to the half-life would give a higher average and a smaller swing, which is the trade-off behind a dosing schedule. Maximum 400 mg, minimum 200 mg

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Write row 4 of Pascal's triangle.
   Show the full solution

   1, 4, 6, 4, 1

2. Evaluate \( \dbinom{5}{2} \).
   Show the full solution

   \( \dfrac{120}{2 \times 6} \). 10

3. Expand \( (x + 1)^3 \).
   Show the full solution

   Row 3 is 1, 3, 3, 1. \( x^3 + 3x^2 + 3x + 1 \)

4. How many terms does \( (a + b)^6 \) have?
   Show the full solution

   Row 6 has 7 entries. 7

5. Evaluate \( \dbinom{6}{0} \).
   Show the full solution

   1

6. Expand \( (x - 3)^4 \).
   Show the full solution

   Treat it as \( (x + (-3))^4 \) with row 4: 1, 4, 6, 4, 1. \( x^4 + 4x^3(-3) + 6x^2(9) + 4x(-27) + 81 \) \( = x^4 - 12x^3 + 54x^2 - 108x + 81 \). The signs alternate because odd powers of \( -3 \) are negative. Check at \( x = 1 \): original gives \( (-2)^4 = 16 \); expansion gives \( 1 - 12 + 54 - 108 + 81 = 16 \) ✓ \( x^4 - 12x^3 + 54x^2 - 108x + 81 \)

7. Find the \( x^2 \) term of \( (x + 5)^6 \).
   Show the full solution

   General term: \( \dbinom{6}{k} x^{6-k} 5^k \). For \( x^2 \): \( 6 - k = 2 \), so \( k = 4 \). \( \dbinom{6}{4} = \dfrac{720}{24 \times 2} = 15 \). \( 5^4 = 625 \). Term \( = 15(625)x^2 = 9375x^2 \). \( 9375x^2 \)

8. Find the constant term of \( (3x + 2)^5 \).
   Show the full solution

   The constant term has \( x^0 \), so \( 5 - k = 0 \) and \( k = 5 \). \( \dbinom{5}{5} = 1 \), \( (3x)^0 = 1 \), \( 2^5 = 32 \). Term \( = 32 \). Check directly. Setting \( x = 0 \) in the original gives \( (0 + 2)^5 = 32 \) ✓ Substituting zero is the fastest check for a constant term, since every other term vanishes. 32

9. Explain why the binomial coefficients count combinations.
   Show the full solution

   Because expanding the product requires choosing one term from each factor, and collecting like terms counts how many choices produce the same result. The setup. \( (a+b)^n \) is \( n \) factors, each offering \( a \) or \( b \). Multiplying out before simplifying produces \( 2^n \) products, one for each pattern of choices. Which products are alike. A product equals \( a^{n-k}b^k \) exactly when \( b \) was chosen from \( k \) of the factors. Which \( k \) does not affect the product, only how many. So the coefficient is a count. It is the number of ways to select \( k \) factors out of \( n \), which is \( \dbinom{n}{k} \) by definition. Verify on \( (a+b)^3 \). There are \( 2^3 = 8 \) products: aaa, aab, aba, baa, abb, bab, bba, bbb. Grouping: one \( a^3 \), three \( a^2b \), three \( ab^2 \), one \( b^3 \). Coefficients 1, 3, 3, 1, matching row 3 ✓ Why the rows are symmetric. Choosing which \( k \) factors give \( b \) is the same as choosing which \( n - k \) give \( a \), so \( \dbinom{n}{k} = \dbinom{n}{n-k} \). The symmetry of Pascal's triangle is that fact. Why row \( n \) sums to \( 2^n \). Every one of the \( 2^n \) products lands in exactly one group, so the group sizes must total \( 2^n \). Row 3 sums to 8 ✓ Substituting \( a = b = 1 \) proves it in one line. Each coefficient counts the ways of choosing which factors contribute the \( b \)

10. Find the term containing \( x^4 \) in \( \left( x^2 + \dfrac{1}{x} \right)^8 \).
    Show the full solution

    Write the general term. \[ \binom{8}{k}\left( x^2 \right)^{8-k}\left( \frac{1}{x} \right)^{k} \] Find the power of \( x \). \( \left( x^2 \right)^{8-k} = x^{16 - 2k} \) and \( \left( \dfrac{1}{x} \right)^k = x^{-k} \). Combining: \( x^{16 - 2k - k} = x^{16 - 3k} \). Set the exponent to 4 and solve. \( 16 - 3k = 4 \), so \( 3k = 12 \) and \( k = 4 \). An integer in range, so the term exists. Had \( k \) come out fractional or outside 0 to 8, there would be no such term, and that is a real possibility worth checking rather than assuming. Compute the coefficient. \( \dbinom{8}{4} = \dfrac{8!}{4!\,4!} = \dfrac{40320}{24 \times 24} = \dfrac{40320}{576} = 70 \). Assemble the term. \( 70 \left( x^2 \right)^4 \left( \dfrac{1}{x} \right)^4 = 70 \cdot x^8 \cdot x^{-4} = 70x^4 \). Verify the exponent arithmetic. With \( k = 4 \): \( 16 - 3(4) = 4 \) ✓ Note what makes this harder than the earlier problems. Both parts of the binomial contain \( x \), so the exponent of \( x \) in a term depends on \( k \) through both pieces. Tracking it as a single linear expression in \( k \) is what keeps it manageable. A related question the same setup answers. Is there a constant term? Set \( 16 - 3k = 0 \), giving \( k = \dfrac{16}{3} \), not an integer. So no constant term exists in this expansion. The method answers existence questions as easily as value questions. \( 70x^4 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Convert \( 90^\circ \) to radians.
   Show the full solution

   \( \frac{\pi}{2} \)

2. Convert \( \pi \) radians to degrees.
   Show the full solution

   \( 180^\circ \)

3. Find a positive angle coterminal with \( -50^\circ \).
   Show the full solution

   \( -50 + 360 \). \( 310^\circ \)

4. Find the arc length for \( r = 5 \), \( \theta = 2 \) radians.
   Show the full solution

   \( s = 5(2) \). 10

5. How many radians are in a full circle?
   Show the full solution

   \( 2\pi \)

6. Convert \( 225^\circ \) to radians and \( \dfrac{7\pi}{6} \) to degrees.
   Show the full solution

   \( 225 \times \dfrac{\pi}{180} = \dfrac{225\pi}{180} = \dfrac{5\pi}{4} \). \( \dfrac{7\pi}{6} \times \dfrac{180}{\pi} = \dfrac{7(180)}{6} = 210^\circ \). Both land in the third quadrant, between \( 180^\circ \) and \( 270^\circ \), consistent with each other's position ✓ \( \frac{5\pi}{4} \) and \( 210^\circ \)

7. Find the sector area for \( r = 6 \), \( \theta = \dfrac{\pi}{4} \).
   Show the full solution

   \( A = \dfrac{1}{2}(36)\left( \dfrac{\pi}{4} \right) = \dfrac{36\pi}{8} = \dfrac{9\pi}{2} \approx 14.14 \). Check: the full circle has area \( 36\pi \approx 113.10 \), and \( \dfrac{\pi}{4} \) is one eighth of a turn. One eighth of 113.10 is 14.14 ✓ \( \frac{9\pi}{2} \approx 14.14 \)

8. A pendulum 30 inches long swings through an angle of \( 40^\circ \). Find the distance the bob travels.
   Show the full solution

   Convert to radians first, since the arc formula requires it. \( 40 \times \dfrac{\pi}{180} = \dfrac{2\pi}{9} \approx 0.6981 \) radians. \( s = r\theta = 30(0.6981) \approx 20.94 \) inches. Check a different way: \( 40^\circ \) is \( \dfrac{1}{9} \) of a full turn, and the full circle of radius 30 has circumference \( 60\pi \approx 188.50 \). One ninth is 20.94 ✓ Using \( 40 \) directly in \( s = r\theta \) would give 1200 inches, a hundred feet, which the physical setup rejects at a glance. About 20.9 inches

9. Explain why radian measure has no units.
   Show the full solution

   Because it is defined as one length divided by another, and the units cancel. The definition. \( \theta = \dfrac{s}{r} \), where \( s \) is an arc length and \( r \) a radius. Both are lengths, so the quotient is a pure number. Check with a concrete case. An arc of 6 cm on a circle of radius 3 cm gives \( \theta = \dfrac{6 \text{ cm}}{3 \text{ cm}} = 2 \). The centimeters cancel, leaving 2, and the answer is the same in inches or meters. Why "radians" is still written. As a reminder of which convention is in use, not as a genuine unit. Degrees, by contrast, genuinely are a unit: a degree is \( \dfrac{1}{360} \) of a turn, a choice with no mathematical content. What being dimensionless buys. A radian measure can be multiplied by a length to give a length, as in \( s = r\theta \), without any conversion factor. A degree measure cannot, which is why the degree version of that formula carries \( \dfrac{\pi}{180} \). Where it matters most. In any formula mixing angles with other physical quantities, such as angular velocity in radians per second, the dimensionless angle lets the units work out correctly on their own. It is an arc length divided by a radius, so the length units cancel

10. A circular pizza of radius 9 inches is cut into 8 equal slices. Find one slice's arc length, area, and the angle in both units, then find what radius would give a slice of exactly 20 square inches at the same angle.
    Show the full solution

    Find the angle. Eight equal slices means each is \( \dfrac{1}{8} \) of a full turn. In degrees: \( \dfrac{360}{8} = 45^\circ \). In radians: \( \dfrac{2\pi}{8} = \dfrac{\pi}{4} \approx 0.7854 \). Find the arc length. \( s = r\theta = 9 \left( \dfrac{\pi}{4} \right) = \dfrac{9\pi}{4} \approx 7.07 \) inches of crust. Check: the full circumference is \( 18\pi \approx 56.55 \), and one eighth is 7.07 ✓ Find the area. \( A = \dfrac{1}{2}r^2\theta = \dfrac{1}{2}(81)\left( \dfrac{\pi}{4} \right) = \dfrac{81\pi}{8} \approx 31.81 \) square inches. Check: the full area is \( 81\pi \approx 254.47 \), and one eighth is 31.81 ✓ Now solve for the radius giving 20 square inches. \( \dfrac{1}{2}r^2\left( \dfrac{\pi}{4} \right) = 20 \). \( \dfrac{\pi r^2}{8} = 20 \), so \( r^2 = \dfrac{160}{\pi} \approx 50.93 \). \( r \approx 7.14 \) inches. Check. \( \dfrac{1}{2}(50.93)\left( \dfrac{\pi}{4} \right) = \dfrac{50.93\pi}{8} \approx 20.00 \) ✓ Note the relationship between the two radii. The ratio of areas is \( \dfrac{20}{31.81} = 0.6287 \), and the ratio of radii is \( \dfrac{7.14}{9} = 0.7935 \). Squaring 0.7935 gives 0.6296, matching to rounding. Area scales as the square of the radius, which is the same relationship Geometry established for similar figures. The practical reading. To cut the slice area by about a third, the pizza's radius shrinks by only about a fifth. Diameter is a poor guide to how much pizza is being bought, and doubling the diameter quadruples the food. Arc \( \frac{9\pi}{4} \approx 7.07 \) in, area \( \frac{81\pi}{8} \approx 31.81 \) in², angle \( 45^\circ = \frac{\pi}{4} \); a radius of about 7.14 in gives 20 in²

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find \( \cos 0 \).
   Show the full solution

   1

2. Find \( \sin \dfrac{\pi}{2} \).
   Show the full solution

   1

3. Find \( \cos \dfrac{\pi}{3} \).
   Show the full solution

   \( \dfrac{\pi}{3} = 60^\circ \). \( \frac{1}{2} \)

4. Find \( \sin \dfrac{\pi}{4} \).
   Show the full solution

   \( \frac{\sqrt{2}}{2} \)

5. What is the largest value \( \sin\theta \) can take?
   Show the full solution

   1

6. Find the coordinates on the unit circle at \( \dfrac{5\pi}{6} \).
   Show the full solution

   \( \dfrac{5\pi}{6} = 150^\circ \), in the second quadrant with reference angle \( 30^\circ \). At \( 30^\circ \) the coordinates are \( \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right) \). In the second quadrant \( x \) is negative and \( y \) is positive: \( \left( -\dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right) \). Check on the circle: \( \dfrac{3}{4} + \dfrac{1}{4} = 1 \) ✓ \( \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right) \)

7. Given \( \sin\theta = \dfrac{3}{5} \) with \( \theta \) in the first quadrant, find \( \cos\theta \).
   Show the full solution

   Use \( \cos^2\theta + \sin^2\theta = 1 \): \( \cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25} \). \( \cos\theta = \pm\dfrac{4}{5} \), and the first quadrant has positive \( x \), so \( \cos\theta = \dfrac{4}{5} \). The quadrant is what selects the sign, and without it both answers would be possible. Check: \( \dfrac{16}{25} + \dfrac{9}{25} = 1 \) ✓ \( \frac{4}{5} \)

8. Find \( \sin \dfrac{7\pi}{4} \) and \( \cos \dfrac{7\pi}{4} \).
   Show the full solution

   \( \dfrac{7\pi}{4} = 315^\circ \), in the fourth quadrant with reference angle \( 45^\circ \). At \( 45^\circ \) both coordinates are \( \dfrac{\sqrt{2}}{2} \). In the fourth quadrant \( x \) is positive and \( y \) is negative. \( \cos \dfrac{7\pi}{4} = \dfrac{\sqrt{2}}{2} \), \( \sin \dfrac{7\pi}{4} = -\dfrac{\sqrt{2}}{2} \). Check: \( \dfrac{1}{2} + \dfrac{1}{2} = 1 \) ✓ \( \sin = -\frac{\sqrt{2}}{2} \), \( \cos = \frac{\sqrt{2}}{2} \)

9. Explain why the unit circle definition works for angles a right triangle cannot represent.
   Show the full solution

   Because it asks only where the terminal side lands, and every angle has a terminal side. The triangle's limitation. A right triangle's angles must total \( 180^\circ \) with one of them \( 90^\circ \), so the other two are strictly between \( 0^\circ \) and \( 90^\circ \). There is no triangle containing a \( 150^\circ \) angle as one of its acute angles, and none at all for \( -40^\circ \) or \( 500^\circ \). What the circle needs instead. An angle in standard position rotates the terminal side from the positive \( x \)-axis. Any amount of rotation is allowed, in either direction, past a full turn or several. The terminal side always meets the circle at exactly one point, and that point's coordinates are the definition. What changes and what does not. The values can now be negative, since coordinates can be. They still satisfy \( \cos^2 + \sin^2 = 1 \), since every point is on the circle. And they repeat every full turn, since coterminal angles share a point. Why the agreement on acute angles matters. Everything proved in Geometry stays true, so the extension costs nothing. A definition that disagreed on the overlap would not be an extension at all. What it makes possible. Periodicity. The repetition every \( 2\pi \) is what lets these functions model anything that cycles, and a triangle-bound definition could not produce it. Every angle has a terminal side meeting the circle once, while only acute angles fit inside a right triangle

10. Find all angles \( \theta \) in \( [0, 2\pi) \) with \( \sin\theta = \dfrac{1}{2} \), and explain why there are exactly that many.
    Show the full solution

    Read the question geometrically. \( \sin\theta \) is the \( y \)-coordinate, so this asks where the unit circle has height \( \dfrac{1}{2} \). Find the intersections. The horizontal line \( y = \dfrac{1}{2} \) cuts the unit circle in two points, one on the right half and one on the left. Find their \( x \)-coordinates. \( x^2 + \dfrac{1}{4} = 1 \), so \( x^2 = \dfrac{3}{4} \) and \( x = \pm\dfrac{\sqrt{3}}{2} \). The points are \( \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right) \) and \( \left( -\dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right) \). Identify the angles. The first is the familiar \( 30^\circ \) point, so \( \theta = \dfrac{\pi}{6} \). The second is its mirror image across the \( y \)-axis, at \( 180^\circ - 30^\circ = 150^\circ \), so \( \theta = \dfrac{5\pi}{6} \). Verify. \( \sin \dfrac{\pi}{6} = \dfrac{1}{2} \) ✓ \( \sin \dfrac{5\pi}{6} = \dfrac{1}{2} \), since the second quadrant keeps \( y \) positive ✓ Why exactly two. A horizontal line at any height strictly between \( -1 \) and 1 cuts a circle in exactly two points. Each point corresponds to one angle in a single revolution, so there are exactly two solutions in \( [0, 2\pi) \). The exceptional heights. At \( y = 1 \) and \( y = -1 \) the line is tangent and meets the circle once, so \( \sin\theta = 1 \) has the single solution \( \dfrac{\pi}{2} \) and \( \sin\theta = -1 \) has only \( \dfrac{3\pi}{2} \). Above 1 or below \( -1 \) there are no solutions at all, which is why \( \sin\theta = 2 \) is impossible. Beyond one revolution. Every coterminal angle also works, so the full solution set is \( \dfrac{\pi}{6} + 2\pi n \) and \( \dfrac{5\pi}{6} + 2\pi n \) for every integer \( n \). Lesson 10.5 makes this general form the standard way to answer. \( \theta = \frac{\pi}{6} \) and \( \frac{5\pi}{6} \); a horizontal line at a height between \( -1 \) and 1 cuts the circle exactly twice

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Which function is the reciprocal of cosine?
   Show the full solution

   Secant

2. Find \( \tan\theta \) if \( \sin\theta = 0.6 \) and \( \cos\theta = 0.8 \).
   Show the full solution

   \( \dfrac{0.6}{0.8} \). 0.75

3. Find \( r \) for the point \( (3, 4) \).
   Show the full solution

   5

4. Find \( \csc\theta \) if \( \sin\theta = \dfrac{2}{7} \).
   Show the full solution

   \( \frac{7}{2} \)

5. Where is \( \tan\theta \) undefined in \( [0, 2\pi) \)?
   Show the full solution

   Where \( \cos\theta = 0 \). \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \)

6. The terminal side passes through \( (5, -12) \). Find all six functions.
   Show the full solution

   \( r = \sqrt{25 + 144} = \sqrt{169} = 13 \). \( \sin = -\dfrac{12}{13} \), \( \cos = \dfrac{5}{13} \), \( \tan = -\dfrac{12}{5} \). \( \csc = -\dfrac{13}{12} \), \( \sec = \dfrac{13}{5} \), \( \cot = -\dfrac{5}{12} \). The point is in the fourth quadrant, where cosine is positive and the other two primaries are negative. Consistent ✓ Check: \( \left( \dfrac{5}{13} \right)^2 + \left( -\dfrac{12}{13} \right)^2 = \dfrac{25 + 144}{169} = 1 \) ✓ \( \sin = -\frac{12}{13} \), \( \cos = \frac{5}{13} \), \( \tan = -\frac{12}{5} \), \( \csc = -\frac{13}{12} \), \( \sec = \frac{13}{5} \), \( \cot = -\frac{5}{12} \)

7. Given \( \cos\theta = -\dfrac{8}{17} \) with \( \theta \) in the third quadrant, find the other five functions.
   Show the full solution

   Use \( \sin^2\theta = 1 - \cos^2\theta = 1 - \dfrac{64}{289} = \dfrac{225}{289} \). \( \sin\theta = \pm\dfrac{15}{17} \), and the third quadrant has negative \( y \), so \( \sin\theta = -\dfrac{15}{17} \). \( \tan\theta = \dfrac{-15/17}{-8/17} = \dfrac{15}{8} \), positive, as the third quadrant requires. \( \csc\theta = -\dfrac{17}{15} \), \( \sec\theta = -\dfrac{17}{8} \), \( \cot\theta = \dfrac{8}{15} \). The quadrant is what chose the sine's sign, and the tangent's sign then followed without a separate decision. \( \sin = -\frac{15}{17} \), \( \tan = \frac{15}{8} \), \( \csc = -\frac{17}{15} \), \( \sec = -\frac{17}{8} \), \( \cot = \frac{8}{15} \)

8. Find all six functions at \( \theta = \dfrac{3\pi}{4} \).
   Show the full solution

   \( \dfrac{3\pi}{4} = 135^\circ \), second quadrant, reference angle \( 45^\circ \). Unit circle point: \( \left( -\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right) \). \( \sin = \dfrac{\sqrt{2}}{2} \), \( \cos = -\dfrac{\sqrt{2}}{2} \), \( \tan = \dfrac{\sqrt{2}/2}{-\sqrt{2}/2} = -1 \). \( \csc = \dfrac{2}{\sqrt{2}} = \sqrt{2} \), \( \sec = -\sqrt{2} \), \( \cot = -1 \). Tangent is \( -1 \) because the terminal side is the line \( y = -x \), whose slope is \( -1 \). That is the geometric reading from the worked example. \( \sin = \frac{\sqrt{2}}{2} \), \( \cos = -\frac{\sqrt{2}}{2} \), \( \tan = -1 \), \( \csc = \sqrt{2} \), \( \sec = -\sqrt{2} \), \( \cot = -1 \)

9. Explain why any point on the terminal side gives the same function values.
   Show the full solution

   Because the right triangles formed by different points on the same ray are similar, and similar triangles have equal ratios of corresponding sides. The setup. Take two points \( P \) and \( Q \) on the same terminal side, and drop perpendiculars from each to the \( x \)-axis. Both triangles have a right angle and share the angle \( \theta \) at the origin, so they are similar by angle angle. What similarity gives. Corresponding sides are proportional. If \( Q \) is \( k \) times as far out as \( P \), then its coordinates are \( kx \) and \( ky \), and its distance is \( kr \). The ratios are unchanged. \( \dfrac{ky}{kr} = \dfrac{y}{r} \), so the sine is the same. Every other function is also a ratio of two of these three quantities, so the \( k \) cancels in each. A concrete check. \( (3, 4) \) gives \( \sin = \dfrac{4}{5} \). \( (30, 40) \) has \( r = 50 \) and gives \( \dfrac{40}{50} = \dfrac{4}{5} \) ✓ Why this matters for the definition. A function must give one output per input. If different points gave different values, "the sine of \( \theta \)" would be meaningless. Similarity is what makes the definition legitimate. Why the unit circle is the convenient choice. Taking \( r = 1 \) makes the denominators disappear, so the coordinates are the cosine and sine. Any radius would work; radius 1 is just the least cluttered. Different points give similar triangles, whose corresponding ratios are equal

10. Given \( \tan\theta = -\dfrac{3}{4} \), find \( \sin\theta \) and \( \cos\theta \) for both possible quadrants, and explain why two answers exist.
    Show the full solution

    Identify the possible quadrants. Tangent is \( \dfrac{y}{x} \), so a negative tangent means \( x \) and \( y \) have opposite signs. That happens in quadrant II, where \( x \lt 0 \) and \( y \gt 0 \), and in quadrant IV, where \( x \gt 0 \) and \( y \lt 0 \). Find \( r \) in each case. In quadrant II, take \( (x, y) = (-4, 3) \): \( r = \sqrt{16 + 9} = 5 \). In quadrant IV, take \( (x, y) = (4, -3) \): \( r = 5 \). Both give \( \tan\theta = -\dfrac{3}{4} \) ✓ Quadrant II answer. \( \sin\theta = \dfrac{3}{5} \), \( \cos\theta = -\dfrac{4}{5} \). Quadrant IV answer. \( \sin\theta = -\dfrac{3}{5} \), \( \cos\theta = \dfrac{4}{5} \). Verify both. Quadrant II: \( \dfrac{9}{25} + \dfrac{16}{25} = 1 \) ✓ and \( \dfrac{3/5}{-4/5} = -\dfrac{3}{4} \) ✓ Quadrant IV: same sum ✓ and \( \dfrac{-3/5}{4/5} = -\dfrac{3}{4} \) ✓ Why two answers exist. Knowing the tangent fixes the slope of the terminal side, and a line through the origin with a given slope extends in two opposite directions. Each direction is a different angle, and the two differ by \( \pi \). What that says about tangent's period. Since opposite directions give the same tangent, \( \tan(\theta + \pi) = \tan\theta \). Tangent repeats every \( \pi \), not every \( 2\pi \) as sine and cosine do. Sine and cosine distinguish the two directions because they depend on the actual coordinates, not just their ratio. What extra information would settle it. Any one of: the quadrant, the sign of sine, or the sign of cosine. Problems that give only the tangent and expect one answer are underdetermined, and noticing that is the point of the exercise. Quadrant II: \( \sin = \frac{3}{5} \), \( \cos = -\frac{4}{5} \). Quadrant IV: \( \sin = -\frac{3}{5} \), \( \cos = \frac{4}{5} \). Tangent fixes a slope, and a line has two directions

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the reference angle for \( 150^\circ \).
   Show the full solution

   \( 180 - 150 \). \( 30^\circ \)

2. In which quadrants is sine positive?
   Show the full solution

   I and II

3. Find the reference angle for \( 240^\circ \).
   Show the full solution

   \( 240 - 180 \). \( 60^\circ \)

4. Is \( \cos 200^\circ \) positive or negative?
   Show the full solution

   Quadrant III has \( x \lt 0 \). Negative

5. In which quadrants is tangent positive?
   Show the full solution

   I and III

6. Evaluate \( \cos 225^\circ \) exactly.
   Show the full solution

   Quadrant III, reference angle \( 225 - 180 = 45^\circ \). At \( 45^\circ \), \( \cos = \dfrac{\sqrt{2}}{2} \). Quadrant III has \( x \lt 0 \), so the value is negative. \( -\frac{\sqrt{2}}{2} \)

7. Evaluate \( \tan \dfrac{4\pi}{3} \) exactly.
   Show the full solution

   \( \dfrac{4\pi}{3} = 240^\circ \), quadrant III, reference angle \( 60^\circ \). At \( 60^\circ \), \( \tan = \sqrt{3} \). Quadrant III has both coordinates negative, so tangent is positive. \( \sqrt{3} \)

8. Evaluate \( \sin(-120^\circ) \) exactly.
   Show the full solution

   Add \( 360^\circ \): coterminal with \( 240^\circ \), quadrant III, reference angle \( 60^\circ \). At \( 60^\circ \), \( \sin = \dfrac{\sqrt{3}}{2} \). Quadrant III has \( y \lt 0 \), so the value is negative. Faster route. \( \sin(-\theta) = -\sin\theta \), and \( \sin 120^\circ = \dfrac{\sqrt{3}}{2} \), so the answer is \( -\dfrac{\sqrt{3}}{2} \). Both routes agree ✓ \( -\frac{\sqrt{3}}{2} \)

9. Explain why the sign pattern can be derived rather than memorized.
   Show the full solution

   Because each function is a ratio built from \( x \), \( y \) and \( r \), and only \( x \) and \( y \) change sign. The one fact that does need knowing. Where the coordinates are positive: \( x \) is positive to the right of the \( y \)-axis, \( y \) is positive above the \( x \)-axis. That is ordinary coordinate geometry, not trigonometry. The second fact. \( r \) is a distance, so it is always positive and never affects a sign. Deriving each function's pattern. Sine is \( \dfrac{y}{r} \), so it has the sign of \( y \): positive in I and II. Cosine is \( \dfrac{x}{r} \), so it has the sign of \( x \): positive in I and IV. Tangent is \( \dfrac{y}{x} \), positive when the signs agree: I and III. The reciprocals follow free. A reciprocal has the same sign as what it inverts, so cosecant matches sine, secant matches cosine, cotangent matches tangent. Why this is more reliable than a mnemonic. A mnemonic gives no way to check itself. Someone who misremembers it has no recourse. The coordinate argument can be redone in fifteen seconds and will always come out right, and it also explains the answer rather than just producing it. A place the mnemonic is silent. It names which functions are positive but says nothing about the reciprocals or about why. The coordinate reasoning covers all six at once. Sine follows \( y \), cosine follows \( x \), tangent follows their quotient, and \( r \) is always positive

10. Find all \( \theta \) in \( [0^\circ, 360^\circ) \) with \( \cos\theta = -\dfrac{1}{2} \), and state the general solution.
    Show the full solution

    Find the reference angle. Ignore the sign for a moment: \( \cos(\text{ref}) = \dfrac{1}{2} \) gives \( \text{ref} = 60^\circ \), one of the known values. Determine the quadrants. Cosine is negative where \( x \lt 0 \), which is quadrants II and III. Convert the reference angle into each quadrant. Quadrant II: \( 180 - 60 = 120^\circ \). Quadrant III: \( 180 + 60 = 240^\circ \). Verify both. \( \cos 120^\circ \): reference \( 60^\circ \), quadrant II, \( x \lt 0 \), so \( -\dfrac{1}{2} \) ✓ \( \cos 240^\circ \): reference \( 60^\circ \), quadrant III, \( x \lt 0 \), so \( -\dfrac{1}{2} \) ✓ Check geometrically. The vertical line \( x = -\dfrac{1}{2} \) cuts the unit circle at two points, one above and one below the \( x \)-axis. Two solutions, as found. State the general solution. Adding any whole number of revolutions gives another solution, so \( \theta = 120^\circ + 360^\circ n \) or \( \theta = 240^\circ + 360^\circ n \) for every integer \( n \). In radians: \( \dfrac{2\pi}{3} + 2\pi n \) and \( \dfrac{4\pi}{3} + 2\pi n \). A symmetry worth noticing. The two solutions are \( 120^\circ \) and \( 240^\circ \), which average to \( 180^\circ \). Cosine solutions always pair symmetrically about \( 0^\circ \) or \( 180^\circ \), because cosine is determined by the horizontal position and the circle is symmetric across the \( x \)-axis. Sine solutions pair about \( 90^\circ \) instead, for the corresponding reason. Using that as a check. Having found \( 120^\circ \), the second solution is \( 360 - 120 = 240^\circ \), found without any quadrant reasoning at all. \( 120^\circ \) and \( 240^\circ \); in general \( 120^\circ + 360^\circ n \) and \( 240^\circ + 360^\circ n \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Give the amplitude of \( y = 5\sin x \).
   Show the full solution

   5

2. Give the period of \( y = \sin(3x) \).
   Show the full solution

   \( \dfrac{2\pi}{3} \). \( \frac{2\pi}{3} \)

3. Give the amplitude of \( y = -4\cos x \).
   Show the full solution

   Amplitude is a distance. 4

4. What is the period of \( y = \cos x \)?
   Show the full solution

   \( 2\pi \)

5. At what \( x \) does \( y = \cos x \) first equal \( -1 \)?
   Show the full solution

   \( \pi \)

6. Give the amplitude and period of \( y = 6\sin\left( \dfrac{x}{2} \right) \).
   Show the full solution

   \( a = 6 \), so amplitude 6. \( b = \dfrac{1}{2} \), so period \( \dfrac{2\pi}{1/2} = 4\pi \). Check: at \( x = \pi \), \( 6\sin\left( \dfrac{\pi}{2} \right) = 6 \), the maximum. A quarter of \( 4\pi \) is \( \pi \), so the first peak should be there ✓ Amplitude 6, period \( 4\pi \)

7. Give five key points for \( y = 2\cos(4x) \).
   Show the full solution

   Amplitude 2, period \( \dfrac{2\pi}{4} = \dfrac{\pi}{2} \). Quarter points: \( 0, \dfrac{\pi}{8}, \dfrac{\pi}{4}, \dfrac{3\pi}{8}, \dfrac{\pi}{2} \). Cosine starts at its maximum: \( (0, 2) \), \( \left( \dfrac{\pi}{8}, 0 \right) \), \( \left( \dfrac{\pi}{4}, -2 \right) \), \( \left( \dfrac{3\pi}{8}, 0 \right) \), \( \left( \dfrac{\pi}{2}, 2 \right) \). Check at \( x = \dfrac{\pi}{4} \): \( 2\cos(\pi) = -2 \) ✓ \( (0,2) \), \( \left( \frac{\pi}{8},0 \right) \), \( \left( \frac{\pi}{4},-2 \right) \), \( \left( \frac{3\pi}{8},0 \right) \), \( \left( \frac{\pi}{2},2 \right) \)

8. Write an equation for a sine curve with amplitude 7 and period \( \dfrac{\pi}{3} \).
   Show the full solution

   Amplitude 7 gives \( a = 7 \). Period \( \dfrac{2\pi}{b} = \dfrac{\pi}{3} \), so \( b = \dfrac{2\pi}{\pi/3} = 2\pi \cdot \dfrac{3}{\pi} = 6 \). \( y = 7\sin(6x) \). Check the period: \( \dfrac{2\pi}{6} = \dfrac{\pi}{3} \) ✓ Solving for \( b \) from the period is the reverse of reading it off, and it is the direction every modeling problem needs. \( y = 7\sin(6x) \)

9. Explain why sine and cosine are the same curve shifted, and give the shift.
   Show the full solution

   They track the two coordinates of the same rotating point, and those coordinates run through the identical pattern a quarter turn apart. The observation. \( \cos x \) starts at 1 and falls to 0 at \( \dfrac{\pi}{2} \). \( \sin x \) starts at 0 and rises to 1 at \( \dfrac{\pi}{2} \). The cosine reaches each of its landmarks a quarter period before the sine reaches the corresponding one. The shift stated. \( \cos x = \sin\left( x + \dfrac{\pi}{2} \right) \). Cosine is sine shifted left by \( \dfrac{\pi}{2} \). Equivalently, \( \sin x = \cos\left( x - \dfrac{\pi}{2} \right) \). Verify at several points. At \( x = 0 \): \( \cos 0 = 1 \) and \( \sin\left( \dfrac{\pi}{2} \right) = 1 \) ✓ At \( x = \dfrac{\pi}{2} \): \( \cos\left( \dfrac{\pi}{2} \right) = 0 \) and \( \sin(\pi) = 0 \) ✓ At \( x = \pi \): \( \cos\pi = -1 \) and \( \sin\left( \dfrac{3\pi}{2} \right) = -1 \) ✓ Why the unit circle makes it obvious. Cosine is the horizontal coordinate and sine the vertical. Rotating the whole picture by \( 90^\circ \) turns one axis into the other, so one function becomes the other with a quarter-turn offset. What follows practically. Any sinusoid can be written with either function by adjusting the phase shift. Lesson 9.7 uses that: whichever function starts where the data starts is chosen, purely to avoid a shift. Cosine is sine shifted left by \( \frac{\pi}{2} \)

10. A sinusoid has a maximum of 9 at \( x = 0 \), a minimum of \( -3 \), and a period of 8. Find its equation and verify three points.
    Show the full solution

    Find the midline. It is halfway between the extremes: \( \dfrac{9 + (-3)}{2} = 3 \). So the graph oscillates about \( y = 3 \), and the equation will need \( +3 \) at the end. Find the amplitude. The distance from the midline to a peak: \( 9 - 3 = 6 \). Confirm from below: \( 3 - (-3) = 6 \) ✓ Find \( b \) from the period. \( \dfrac{2\pi}{b} = 8 \), so \( b = \dfrac{2\pi}{8} = \dfrac{\pi}{4} \). Choose the function. The maximum is at \( x = 0 \), and cosine starts at its maximum. So cosine with no horizontal shift. \[ y = 6\cos\left( \frac{\pi}{4}x \right) + 3 \] Verify at \( x = 0 \). \( 6\cos 0 + 3 = 6 + 3 = 9 \) ✓ the stated maximum. Verify at \( x = 4 \), half a period along, where the minimum should be. \( 6\cos\left( \dfrac{\pi}{4} \cdot 4 \right) + 3 = 6\cos\pi + 3 = -6 + 3 = -3 \) ✓ Verify at \( x = 2 \), a quarter period along, where the midline should be crossed. \( 6\cos\left( \dfrac{\pi}{2} \right) + 3 = 0 + 3 = 3 \) ✓ Check the period independently. At \( x = 8 \): \( 6\cos(2\pi) + 3 = 9 \), back to the maximum after exactly 8 units ✓ Could sine have been used instead? Yes, with a shift: \( y = 6\sin\left( \dfrac{\pi}{4}(x + 2) \right) + 3 \) gives the same curve, since shifting sine left a quarter period produces cosine. Checking at \( x = 0 \): \( 6\sin\left( \dfrac{\pi}{2} \right) + 3 = 9 \) ✓ Why cosine was the better choice here. It needed no phase shift, which removes the step where errors most often occur. The rule of thumb: use cosine when the data starts at an extreme, sine when it starts at the midline. \( y = 6\cos\left( \frac{\pi}{4}x \right) + 3 \)