AP Calculus AB

Lesson 1.1 · Unit 1 · CED topic 1.1

Can change occur at an instant?

You already know how to measure change over an interval: pick two points and find the slope between them. Calculus asks a stranger question: how fast is something changing right now, at a single instant, when there's no interval at all?

To get at an instant, shrink the interval. The trouble is that a zero-width interval gives \(\frac{0}{0}\), which is meaningless. The way out is to watch what the average rate approaches as the interval shrinks, and that idea is a limit.

Lesson 1.2 · Unit 1 · CED topic 1.2

Defining limits and limit notation

A limit describes where a function is heading, not where it is. That distinction is the whole point: the value of the function at the target, or whether it even has one, is irrelevant.

Limits also come in one-sided versions. \(\displaystyle\lim_{x \to c^-} f(x)\) approaches from the left, \(\displaystyle\lim_{x \to c^+} f(x)\) from the right. The two-sided limit exists only when both one-sided limits exist and agree.

Lesson 1.3 · Unit 1 · CED topics 1.3–1.4

Estimating limits from graphs and tables

Before you can compute limits algebraically, you need to be able to read them. On the exam, you'll be handed graphs and tables and asked what a function is approaching.

From a graph: trace the curve toward \(x = c\) from each side and read off the \(y\)-value the branches approach. An open circle at \(x = c\) means the function is undefined there: the limit is still the height the branches point to. A filled dot somewhere else is the actual value \(f(c)\); it doesn't affect the limit.

Lesson 1.4 · Unit 1 · CED topics 1.5–1.7

Limit laws and algebraic manipulation

Most limits you'll meet can be found exactly, with algebra. The strategy has two steps: try direct substitution, and if that gives the indeterminate form \(\frac{0}{0}\), rewrite the expression until substitution works.

Lesson 1.5 · Unit 1 · CED topic 1.8

The Squeeze Theorem

Some functions are too wild to handle directly: they oscillate infinitely often near the target. The Squeeze Theorem lets you trap such a function between two tame ones and let the tame ones do the work.

On the exam, the squeeze is usually set up for you: a problem will state inequalities and ask for a limit. Your job is to recognize the theorem and state that the bounds share a limit.

Lesson 1.6 · Unit 1 · CED topics 1.10–1.13

Continuity and types of discontinuities

Informally, a function is continuous if you can draw it without lifting your pencil. The formal version is a three-part checklist, and the exam loves to test each part separately.

When continuity fails, name the failure:

• Removable: the limit exists, but \(f(c)\) is missing or has the wrong value. A hole.
• Jump: the one-sided limits exist but differ. Two branches at different heights.
• Infinite: at least one one-sided limit is \(\pm\infty\). A vertical asymptote.

Lesson 1.7 · Unit 1 · CED topics 1.14–1.15

Infinite limits and asymptotes

Two kinds of "infinity" show up in limits. A function can blow up to \(\pm\infty\) near a finite \(x\), a vertical asymptote, or you can ask where a function settles as \(x\) itself heads to \(\pm\infty\): a horizontal asymptote.

Lesson 1.8 · Unit 1 · CED topic 1.16

The Intermediate Value Theorem

Continuity has a payoff: if a continuous function is below a value at one point and above it at another, it must have hit that value somewhere in between. You can't cross a river without getting wet.

The most common use is proving an equation has a solution without solving it. The exam grades the justification, so learn the sentence structure below word for word.

Three things must appear for full credit: the function is continuous on the closed interval, the target value lies between the endpoint values, and the conclusion names the open interval.

Unit 1 practice · 10 problems

Unit 1 practice: Limits and Continuity

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer: the reveal shows the key steps, not just the number. No calculator needed.

1. Evaluate \(\displaystyle\lim_{x \to 4}\frac{x^2 - 16}{x - 4}\).

   Show answer

   Factor: \(\dfrac{(x-4)(x+4)}{x-4} = x + 4\) for \(x \ne 4\). The limit is \(8\).

2. Evaluate \(\displaystyle\lim_{x \to 0}\frac{\sin 4x}{3x}\).

   Show answer

   \(\dfrac{\sin 4x}{3x} = \dfrac{4}{3}\cdot\dfrac{\sin 4x}{4x} \to \dfrac{4}{3}\cdot 1 = \dfrac{4}{3}\).

3. Evaluate \(\displaystyle\lim_{x \to \infty}\frac{5x^2 - 3x + 1}{2x^2 + 7}\).

   Show answer

   Equal degrees, so the limit is the ratio of leading coefficients: \(\dfrac{5}{2}\).

4. Evaluate \(\displaystyle\lim_{x \to 3^-}\frac{x + 1}{x - 3}\).

   Show answer

   The numerator approaches 4 while the denominator approaches 0 through negative values. The limit is \(-\infty\).

5. Find \(k\) so that \(f(x) = \begin{cases} x^2 + 1 & x < 2 \\ 3x - k & x \ge 2 \end{cases}\) is continuous at \(x = 2\).

   Show answer

   Left limit \(= 5\); right side \(= 6 - k\). Set \(6 - k = 5\), so \(k = 1\).

6. Evaluate \(\displaystyle\lim_{x \to 0}\frac{\sqrt{x + 9} - 3}{x}\).

   Show answer

   Multiply by the conjugate: \(\dfrac{(x + 9) - 9}{x\left(\sqrt{x+9} + 3\right)} = \dfrac{1}{\sqrt{x+9} + 3} \to \dfrac{1}{6}\).

7. Let \(g(x) = \dfrac{x^2 - 5x + 6}{x - 2}\). Classify the discontinuity at \(x = 2\), and give the value of \(g(2)\) that would remove it.

   Show answer

   \(g(x) = \dfrac{(x-2)(x-3)}{x-2} = x - 3\) for \(x \ne 2\), so the limit at 2 is \(-1\). The discontinuity is removable; defining \(g(2) = -1\) removes it.

8. Evaluate \(\displaystyle\lim_{x \to -\infty}\frac{3x + 2}{\sqrt{x^2 + 1}}\).

   Show answer

   For large negative \(x\), \(\sqrt{x^2 + 1} \approx |x| = -x\), so the expression behaves like \(\dfrac{3x}{-x} = -3\). The limit is \(-3\).

9. \(f\) is continuous on \([1, 5]\) with \(f(1) = -2\) and \(f(5) = 7\). Must there be a \(c\) in \((1, 5)\) with \(f(c) = 3\)? Justify.

   Show answer

   Yes. \(f\) is continuous on \([1, 5]\) and \(f(1) = -2 < 3 < 7 = f(5)\), so by the Intermediate Value Theorem there is a \(c\) in \((1, 5)\) with \(f(c) = 3\).

10. Use the Squeeze Theorem to evaluate \(\displaystyle\lim_{x \to 0} x^4\cos\!\left(\frac{2}{x}\right)\).

    Show answer

    \(-x^4 \le x^4\cos\!\left(\tfrac{2}{x}\right) \le x^4\), and both bounds approach 0. The limit is \(0\).

Lesson 2.1 · Unit 2 · CED topic 2.1

Average and instantaneous rates of change at a point

Unit 1 ended with a promise: the limit idea would let us pin down a rate of change at a single instant. Here's the payoff. The average rate over a shrinking interval has a limit, and that limit is the instantaneous rate.

Notice that the algebra did the work the table did in Lesson 1.1. Instead of guessing where the averages were heading, we canceled the \(h\) and read the limit off exactly.

Lesson 2.2 · Unit 2 · CED topic 2.2

Defining the derivative and derivative notation

Do the instantaneous-rate calculation at a general \(x\) instead of a specific number and you get a new function: one that reports the slope of \(f\) everywhere at once. That function is the derivative.

Lesson 2.3 · Unit 2 · CED topic 2.3

Estimating derivatives from tables and graphs

Real data doesn't come with a formula. When all you have is a table or a graph, you estimate the derivative with the best difference quotient available, and the exam expects you to know which one that is.

From a graph: sketch the tangent line and estimate its slope with two points on it. Sign is often enough: the derivative is positive where the graph rises, negative where it falls, and zero at the tops and bottoms of hills. On the calculator sections, your calculator's numerical derivative does this for you when a formula is given.

Lesson 2.4 · Unit 2 · CED topic 2.4

Connecting differentiability and continuity

Every differentiable function is continuous: you can't have a tangent line at a gap. But continuity is not enough for a derivative. A graph can be unbroken and still have no well-defined slope at a point.

Lesson 2.5 · Unit 2 · CED topics 2.5–2.6

The power rule and the basic derivative rules

Computing every derivative from the limit definition would be unbearable. Fortunately a handful of rules, all provable from the definition, handle every polynomial and most algebraic functions.

Lesson 2.6 · Unit 2 · CED topic 2.7

Derivatives of sin x, cos x, eˣ, and ln x

Four derivatives you need cold. Each can be proved from the limit definition: the sine proof uses the two special limits from Lesson 1.4, but on the exam you simply need to know them.

The fact that \(e^x\) is its own derivative is why \(e\) is the natural base: it's the one exponential whose growth rate equals its current value. You'll lean on that heavily in Unit 7.

Lesson 2.7 · Unit 2 · CED topics 2.8–2.9

The product and quotient rules

The derivative of a sum is the sum of the derivatives. The derivative of a product is not the product of the derivatives, and this is the single most common algebra error in Unit 2.

Lesson 2.8 · Unit 2 · CED topic 2.10

Derivatives of tan, cot, sec, and csc

The remaining four trig derivatives all follow from the quotient rule plus \(\sin\) and \(\cos\). Derive one so you trust them, then memorize all four.

Unit 2 practice · 10 problems

Unit 2 practice, Differentiation: Definition and Fundamental Properties

Ten problems covering the whole unit. Work each one on paper before revealing the answer. No calculator needed.

1. Use the limit definition of the derivative to find \(f'(x)\) for \(f(x) = 3x^2 - x\).

   Show answer

   \(\dfrac{3(x+h)^2 - (x+h) - 3x^2 + x}{h} = \dfrac{6xh + 3h^2 - h}{h} = 6x + 3h - 1 \to 6x - 1\).

2. Evaluate \(\displaystyle\lim_{h \to 0}\frac{\sin\!\left(\frac{\pi}{6} + h\right) - \sin\frac{\pi}{6}}{h}\).

   Show answer

   This is the derivative of \(\sin x\) at \(x = \tfrac{\pi}{6}\): \(\cos\tfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}\).

3. Find \(f'(x)\) for \(f(x) = 2x^5 - 4x^3 + 7x - 12\).

   Show answer

   \(f'(x) = 10x^4 - 12x^2 + 7\).

4. Find \(g'(x)\) for \(g(x) = \dfrac{5}{x^3} - 2\sqrt{x} + e^x\).

   Show answer

   Rewrite as \(5x^{-3} - 2x^{1/2} + e^x\). Then \(g'(x) = -15x^{-4} - x^{-1/2} + e^x = -\dfrac{15}{x^4} - \dfrac{1}{\sqrt{x}} + e^x\).

5. Find the equation of the tangent line to \(y = x^2\ln x\) at \(x = 1\).

   Show answer

   \(y(1) = 0\). \(y' = 2x\ln x + x\), so \(y'(1) = 1\). Tangent: \(y = x - 1\).

6. Find \(h'(x)\) for \(h(x) = \dfrac{x^2 + 1}{x - 2}\).

   Show answer

   \(h'(x) = \dfrac{2x(x - 2) - (x^2 + 1)}{(x - 2)^2} = \dfrac{x^2 - 4x - 1}{(x - 2)^2}\).

7. \(f\) is differentiable, with \(f(1) = 3\), \(f(2) = 7\), \(f(3) = 13\), \(f(4) = 21\). Estimate \(f'(3)\).

   Show answer

   Use the points straddling \(x = 3\): \(f'(3) \approx \dfrac{21 - 7}{4 - 2} = 7\).

8. Given \(f(1) = 4\), \(f'(1) = -2\), \(g(1) = 3\), \(g'(1) = 5\), find \((fg)'(1)\) and \(\left(\dfrac{f}{g}\right)'(1)\).

   Show answer

   \((fg)'(1) = (-2)(3) + (4)(5) = 14\). \(\left(\dfrac{f}{g}\right)'(1) = \dfrac{(-2)(3) - (4)(5)}{3^2} = -\dfrac{26}{9}\).

9. Is \(f(x) = \begin{cases} x^2 + 2x & x \le 1 \\ 4x - 1 & x > 1 \end{cases}\) differentiable at \(x = 1\)? Justify.

   Show answer

   Continuity: both pieces equal 3 at \(x = 1\). Slopes: \(2x + 2 = 4\) from the left, \(4\) from the right. Both match, so yes, \(f\) is differentiable at 1 with \(f'(1) = 4\).

10. Find \(f'\!\left(\dfrac{\pi}{3}\right)\) for \(f(x) = \sec x\tan x\).

    Show answer

    \(f'(x) = \sec x\tan^2 x + \sec^3 x\). At \(\tfrac{\pi}{3}\), \(\sec = 2\) and \(\tan = \sqrt{3}\): \(2\cdot 3 + 8 = 14\).

Lesson 3.1 · Unit 3 · CED topic 3.1

The chain rule

Most functions worth differentiating are compositions: a function inside another function. The chain rule says: differentiate the outside (leaving the inside alone), then multiply by the derivative of the inside.

Lesson 3.2 · Unit 3 · CED topic 3.2

Implicit differentiation

Not every curve is a function \(y = f(x)\). A circle isn't. But you can still find slopes on it: differentiate the whole equation with respect to \(x\), treating \(y\) as a function of \(x\) and applying the chain rule every time \(y\) appears.

Horizontal tangents occur where the numerator of \(y'\) is zero (and the denominator isn't); vertical tangents where the denominator is zero (and the numerator isn't).

Lesson 3.3 · Unit 3 · CED topic 3.3

Differentiating inverse functions

The graph of \(f^{-1}\) is the graph of \(f\) reflected over \(y = x\). Reflecting a line with slope \(m\) gives a line with slope \(\frac{1}{m}\), so the derivative of the inverse is the reciprocal of the derivative of the original, evaluated at the matching point.

Lesson 3.4 · Unit 3 · CED topic 3.4

Differentiating inverse trigonometric functions

Implicit differentiation gives the derivatives of the inverse trig functions in a few lines. Watch the derivation for \(\arctan x\), then memorize the three you need most.

Lesson 3.5 · Unit 3 · CED topic 3.5

Selecting procedures for calculating derivatives

You now have every derivative rule in the course. The remaining skill is diagnosis: look at a function, identify its outermost structure, and pick the rule that matches. Simplify before differentiating whenever you can: it prevents most errors.

Lesson 3.6 · Unit 3 · CED topic 3.6

Higher-order derivatives

The derivative of a function is a function, so it has a derivative of its own. The second derivative \(f''\) measures how fast the slope is changing: in motion, that's acceleration. In Unit 5 it will tell you which way a graph bends.

Unit 3 practice · 10 problems

Unit 3 practice: Composite, Implicit, and Inverse Functions

Ten problems covering the whole unit. Work each one on paper before revealing the answer. No calculator needed.

1. Differentiate \(y = (2x^3 - 5)^6\).

   Show answer

   \(y' = 6(2x^3 - 5)^5\cdot 6x^2 = 36x^2(2x^3 - 5)^5\).

2. Differentiate \(y = e^{\sin x}\).

   Show answer

   Chain rule with \(u = \sin x\): \(\dfrac{d}{dx}e^{u} = e^{u}\dfrac{du}{dx}\). Here \(\dfrac{du}{dx} = \cos x\), so \(y' = \cos x\, e^{\sin x}\).

3. Differentiate \(y = \ln(\cos x)\).

   Show answer

   \(y' = \dfrac{-\sin x}{\cos x} = -\tan x\).

4. Find \(\dfrac{dy}{dx}\) for \(x^2 y + y^3 = 10\).

   Show answer

   \(2xy + x^2 y' + 3y^2 y' = 0\), so \(y' = -\dfrac{2xy}{x^2 + 3y^2}\).

5. Find the slope of the tangent line to \(x^2 - xy + y^2 = 7\) at the point \((2, 3)\).

   Show answer

   \(2x - y - xy' + 2yy' = 0 \Rightarrow y' = \dfrac{y - 2x}{2y - x}\). At \((2, 3)\): \(\dfrac{3 - 4}{6 - 2} = -\dfrac{1}{4}\).

6. Let \(f(x) = x^3 + 2x + 1\) and \(g = f^{-1}\). Find \(g'(4)\).

   Show answer

   \(f(1) = 4\), so \(g(4) = 1\). \(f'(x) = 3x^2 + 2\), \(f'(1) = 5\). \(g'(4) = \dfrac{1}{5}\).

7. Differentiate \(y = \arctan(3x)\).

   Show answer

   Use \(\dfrac{d}{dx}\arctan u = \dfrac{u'}{1 + u^{2}}\) with \(u = 3x\), so \(u' = 3\): \(y' = \dfrac{3}{1 + (3x)^{2}} = \dfrac{3}{1 + 9x^2}\).

8. Differentiate \(y = \arcsin(x^2)\).

   Show answer

   Use \(\dfrac{d}{dx}\arcsin u = \dfrac{u'}{\sqrt{1 - u^{2}}}\) with \(u = x^{2}\), so \(u' = 2x\): \(y' = \dfrac{2x}{\sqrt{1 - (x^{2})^{2}}} = \dfrac{2x}{\sqrt{1 - x^4}}\).

9. Let \(h(x) = f(g(x))\), with \(g(3) = 2\), \(g'(3) = -4\), \(f'(2) = 6\), and \(f'(3) = 10\). Find \(h'(3)\).

   Show answer

   \(h'(3) = f'(g(3))\cdot g'(3) = f'(2)\cdot(-4) = -24\). (The value \(f'(3) = 10\) is a distractor.)

10. Find \(y'''\) for \(y = \sin(2x)\).

    Show answer

    \(y' = 2\cos 2x\), \(y'' = -4\sin 2x\), \(y''' = -8\cos 2x\).

Lesson 4.1 · Unit 4 · CED topic 4.1

Interpreting the meaning of the derivative in context

A derivative is a rate, and a rate has units. Free-response questions routinely ask you to explain what a number like \(f'(5) = -3\) means "in the context of the problem", and they grade the sentence, not the number.

Lesson 4.2 · Unit 4 · CED topic 4.2

Straight-line motion: position, velocity, acceleration

A particle moving along a line has position \(x(t)\). Its velocity is \(v(t) = x'(t)\) and its acceleration is \(a(t) = v'(t) = x''(t)\). Almost every motion question comes down to the signs of \(v\) and \(a\).

Lesson 4.3 · Unit 4 · CED topic 4.3

Rates of change in applied contexts other than motion

Anything that changes has a rate of change, and that rate is a derivative. Populations, costs, areas, temperatures: the calculus is identical; only the units and the story change.

Lesson 4.4 · Unit 4 · CED topics 4.4–4.5

Related rates

Two quantities are linked by an equation, and both change with time. Knowing how fast one changes tells you how fast the other does. Differentiate the linking equation with respect to \(t\): every variable picks up a \(\frac{d}{dt}\) by the chain rule.

Lesson 4.5 · Unit 4 · CED topic 4.6

Local linearity and linearization

Zoom in far enough on any differentiable curve and it looks like a straight line: its tangent. That means the tangent line is a good stand-in for the function near the point of tangency, and lines are easy to evaluate.

Lesson 4.6 · Unit 4 · CED topic 4.7

L'Hospital's Rule

Back to limits, now with derivatives in hand. When a limit has the indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), you may replace the fraction by the ratio of the derivatives: top and bottom differentiated separately, not with the quotient rule.

Unit 4 practice · 10 problems

Unit 4 practice: Contextual Applications of Differentiation

Ten problems covering the whole unit. Work each one on paper before revealing the answer. A calculator is fine for the final arithmetic in problems 4, 5, and 10.

1. \(W(t)\) is the number of gallons of water in a pool \(t\) hours after noon, and \(W'(3) = -15\). Interpret this with units.

   Show answer

   At 3:00 pm, the amount of water in the pool is decreasing at a rate of 15 gallons per hour.

2. A particle's position is \(x(t) = t^3 - 9t^2 + 24t\) for \(t \ge 0\). When is it at rest?

   Show answer

   \(v(t) = 3t^2 - 18t + 24 = 3(t - 2)(t - 4)\). At rest at \(t = 2\) and \(t = 4\).

3. For the same particle, is it speeding up or slowing down at \(t = 1\)?

   Show answer

   \(v(1) = 9\) and \(a(t) = 6t - 18\), so \(a(1) = -12\). Opposite signs: slowing down.

4. The edge of a cube grows at 2 cm/s. How fast is the volume increasing when the edge is 5 cm?

   Show answer

   \(V = s^3 \Rightarrow \dfrac{dV}{dt} = 3s^2\dfrac{ds}{dt} = 3(25)(2) = 150\) cm³/s.

5. Two cars leave the same point at the same time; one drives north at 60 mph, the other east at 80 mph. How fast is the distance between them increasing after one hour?

   Show answer

   \(d^2 = x^2 + y^2 \Rightarrow d\,d' = x\,x' + y\,y'\). After 1 h: \(x = 80\), \(y = 60\), \(d = 100\). \(d' = \dfrac{80(80) + 60(60)}{100} = 100\) mph.

6. Use a linearization to approximate \(\sqrt[3]{8.06}\).

   Show answer

   \(f(x) = x^{1/3}\) at \(a = 8\): \(f(8) = 2\), \(f'(8) = \tfrac{1}{3}(8)^{-2/3} = \tfrac{1}{12}\). \(L(8.06) = 2 + \tfrac{0.06}{12} = 2.005\).

7. Evaluate \(\displaystyle\lim_{x \to 0}\frac{e^{2x} - 1}{\sin x}\).

   Show answer

   Form \(\frac{0}{0}\). L'Hospital: \(\displaystyle\lim_{x \to 0}\frac{2e^{2x}}{\cos x} = 2\).

8. Evaluate \(\displaystyle\lim_{x \to \infty}\frac{x^3}{e^x}\).

   Show answer

   Form \(\frac{\infty}{\infty}\); apply L'Hospital three times: \(\dfrac{3x^2}{e^x} \to \dfrac{6x}{e^x} \to \dfrac{6}{e^x} \to 0\).

9. Evaluate \(\displaystyle\lim_{x \to 1}\frac{\ln x}{x^2 - 1}\).

   Show answer

   Form \(\frac{0}{0}\). \(\displaystyle\lim_{x \to 1}\frac{1/x}{2x} = \frac{1}{2}\).

10. A 13 ft ladder leans against a wall. The top slides down at 2 ft/s. How fast is the base moving away from the wall when the base is 5 ft from it?

    Show answer

    \(x^2 + y^2 = 169\); when \(x = 5\), \(y = 12\). \(2x\,x' + 2y\,y' = 0 \Rightarrow 5x' + 12(-2) = 0\), so \(x' = \dfrac{24}{5} = 4.8\) ft/s.

Lesson 5.1 · Unit 5 · CED topic 5.1

The Mean Value Theorem

If you drive 120 miles in 2 hours, at some instant your speedometer read exactly 60. That's the Mean Value Theorem: somewhere on a smooth curve, the tangent line is parallel to the secant line through the endpoints.

Lesson 5.2 · Unit 5 · CED topic 5.2

Extreme Value Theorem, extrema, and critical points

Where can a function be biggest or smallest? Only in a few places: at points where the derivative is zero, where it doesn't exist, or at the ends of the interval. Everything in this unit is built on that fact.

Not every critical point is an extremum: \(f(x) = x^3\) has \(f'(0) = 0\) but just flattens out momentarily at the origin and keeps rising.

Lesson 5.3 · Unit 5 · CED topics 5.3–5.4

Increasing and decreasing intervals; the first derivative test

The sign of \(f'\) tells you which way \(f\) is going. Where the sign changes, \(f\) turns around, and that's a local extremum.

Lesson 5.4 · Unit 5 · CED topic 5.5

Absolute extrema on a closed interval: the candidates test

On a closed interval, the absolute maximum and minimum are guaranteed to exist (Extreme Value Theorem) and guaranteed to be at a critical point or an endpoint. So list the candidates, evaluate \(f\) at each, and compare. No sign charts needed.

Lesson 5.5 · Unit 5 · CED topics 5.6–5.7

Concavity, points of inflection, and the second derivative test

The first derivative says whether a graph rises or falls. The second derivative says which way it bends, like a cup holding water (concave up) or spilling it (concave down).

Lesson 5.6 · Unit 5 · CED topics 5.8–5.9

Sketching graphs and connecting f, f′, and f″

The exam's favorite Unit 5 question hands you the graph of \(f'\), not \(f\), and asks about \(f\). Every feature of \(f\) can be read off the derivative's graph if you know the translation.

Lesson 5.7 · Unit 5 · CED topics 5.10–5.11

Optimization

Largest area, least material, shortest distance: optimization problems turn a word problem into a function of one variable and then find its extreme value with the tools of this unit.

Lesson 5.8 · Unit 5 · CED topic 5.12

Exploring behaviors of implicit relations

The tools of this unit (critical points, the second derivative) work on implicitly defined curves too. The difference is that \(y'\) is a fraction in \(x\) and \(y\), so horizontal tangents come from its numerator and vertical tangents from its denominator.

The logic is the same as for functions; the only extra step is substituting the point's \(y\)-coordinate as well as its \(x\).

Unit 5 practice · 10 problems

Unit 5 practice: Analytical Applications of Differentiation

Ten problems covering the whole unit: one of the two heaviest-weighted units on the exam. Work each one on paper before revealing the answer. No calculator needed.

1. Find the absolute maximum and minimum of \(f(x) = x^3 - 12x\) on \([-3, 3]\).

   Show answer

   \(f'(x) = 3x^2 - 12 = 0\) at \(x = \pm 2\). Candidates: \(f(-3) = 9\), \(f(-2) = 16\), \(f(2) = -16\), \(f(3) = -9\). Absolute max 16 at \(x = -2\); absolute min \(-16\) at \(x = 2\).

2. Find the value of \(c\) guaranteed by the Mean Value Theorem for \(f(x) = \sqrt{x}\) on \([1, 9]\).

   Show answer

   Average rate \(= \dfrac{3 - 1}{8} = \dfrac{1}{4}\). \(f'(c) = \dfrac{1}{2\sqrt{c}} = \dfrac{1}{4} \Rightarrow c = 4\).

3. Find the intervals of concavity and the points of inflection of \(g(x) = x^4 - 8x^2 + 3\).

   Show answer

   \(g''(x) = 12x^2 - 16 = 0 \Rightarrow x = \pm\dfrac{2}{\sqrt{3}}\). Concave up for \(|x| > \dfrac{2}{\sqrt{3}}\), concave down on \(\left(-\dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}\right)\); inflection points at \(x = \pm\dfrac{2}{\sqrt{3}} \approx \pm 1.155\).

4. \(f'(x) = x^2(x - 4)(x + 2)\). At which \(x\) does \(f\) have a local maximum? A local minimum?

   Show answer

   \(x^2 \ge 0\) never changes sign. \((x - 4)(x + 2)\) is positive for \(x < -2\), negative on \((-2, 4)\), positive for \(x > 4\). Local max at \(x = -2\) (\(+\) to \(-\)); local min at \(x = 4\) (\(-\) to \(+\)); nothing at \(x = 0\).

5. Use the second derivative test to classify the critical points of \(h(x) = 2x^3 - 3x^2 - 36x + 5\).

   Show answer

   \(h'(x) = 6(x - 3)(x + 2)\); critical points \(3\) and \(-2\). \(h''(x) = 12x - 6\): \(h''(3) = 30 > 0\) ⇒ local min at 3; \(h''(-2) = -30 < 0\) ⇒ local max at \(-2\).

6. A rectangle has its base on the \(x\)-axis and its upper corners on \(y = 12 - x^2\). Find the maximum possible area.

   Show answer

   \(A(x) = 2x(12 - x^2) = 24x - 2x^3\); \(A'(x) = 24 - 6x^2 = 0 \Rightarrow x = 2\); \(A''(2) < 0\). Maximum area \(= 4 \cdot 8 = 32\) (a 4 by 8 rectangle).

7. An open-top box with a square base must hold 32 in³. Find the dimensions that minimize the surface area.

   Show answer

   \(x^2 h = 32 \Rightarrow h = \dfrac{32}{x^2}\). \(S = x^2 + 4xh = x^2 + \dfrac{128}{x}\); \(S' = 2x - \dfrac{128}{x^2} = 0 \Rightarrow x = 4\), \(h = 2\). Minimum surface area \(= 48\) in².

8. \(f\) is twice differentiable on \([0, 6]\), with \(f' > 0\) on \((0, 2)\), \(f' < 0\) on \((2, 6)\), \(f'' < 0\) on \((0, 4)\), and \(f'' > 0\) on \((4, 6)\). Describe the graph of \(f\).

   Show answer

   \(f\) increases to a local maximum at \(x = 2\), then decreases. It is concave down on \((0, 4)\) and concave up on \((4, 6)\), with a point of inflection at \(x = 4\).

9. Find all points on \(x^2 + 4y^2 = 16\) where the tangent line is horizontal, and all points where it is vertical.

   Show answer

   \(2x + 8yy' = 0 \Rightarrow y' = -\dfrac{x}{4y}\). Horizontal where \(x = 0\): \((0, \pm 2)\). Vertical where \(y = 0\): \((\pm 4, 0)\).

10. Find and classify the local extrema of \(f(x) = \dfrac{x}{x^2 + 1}\).

    Show answer

    \(f'(x) = \dfrac{1 - x^2}{(x^2 + 1)^2}\), zero at \(x = \pm 1\); positive on \((-1, 1)\), negative outside. Local minimum \(f(-1) = -\tfrac{1}{2}\); local maximum \(f(1) = \tfrac{1}{2}\).

Lesson 6.1 · Unit 6 · CED topic 6.1

Exploring accumulation of change

Integration begins with a picture, not a formula. If you know how fast something changes, the area under the rate graph is how much it changed. That single idea is the whole of Unit 6.

Lesson 6.2 · Unit 6 · CED topic 6.2

Approximating areas with Riemann sums

Most rate graphs aren't rectangles and triangles. Riemann sums estimate the area with rectangles anyway, many thin ones, and the exam tests four flavors plus a judgment about whether each over- or underestimates.

Lesson 6.3 · Unit 6 · CED topic 6.3

Summation notation and the definite integral

Use more and more rectangles and the Riemann sum approaches the exact area. That limit is the definite integral. The exam mostly asks you to translate between the two notations.

Lesson 6.4 · Unit 6 · CED topics 6.4–6.5

The Fundamental Theorem of Calculus and accumulation functions

Here is the theorem the whole course has been building toward. An accumulation function adds up \(f\) from a fixed start to a moving endpoint \(x\). Its rate of change is just \(f(x)\): integration and differentiation undo each other.

Lesson 6.5 · Unit 6 · CED topic 6.6

Properties of definite integrals

Because a definite integral is a signed area, it obeys the rules areas do: you can split it, reverse it, scale it, and add it. These properties let you compute integrals from given values without any formula for \(f\).

Lesson 6.6 · Unit 6 · CED topic 6.7

The Fundamental Theorem, Part 2: evaluating definite integrals

Part 1 said accumulation functions have derivative \(f\). Part 2 turns that around into a computing tool: to find the exact area under \(f\), find any function whose derivative is \(f\) and subtract its values at the endpoints.

Lesson 6.7 · Unit 6 · CED topic 6.8

Finding antiderivatives and indefinite integrals

FTC Part 2 makes antiderivatives the central skill. Every derivative rule from Units 2–3, read backwards, is an integration rule, plus a constant, since any constant differentiates to zero.

Lesson 6.8 · Unit 6 · CED topic 6.9

Integrating using substitution

Substitution is the chain rule run backwards. If you can spot an inner function whose derivative also appears (up to a constant), rename the inner function \(u\) and the integral collapses to a basic one.

Lesson 6.9 · Unit 6 · CED topics 6.10, 6.14

Long division, completing the square, and selecting a technique

Two algebraic rewrites unlock rational integrands that substitution can't touch. Then, with every AB technique in hand, the remaining skill is choosing the right one quickly.

Unit 6 practice · 10 problems

Unit 6 practice: Integration and Accumulation of Change

Ten problems covering the whole unit. Work each one on paper before revealing the answer. No calculator needed.

1. Compute the left Riemann sum for \(f(x) = x^2 + 1\) on \([0, 4]\) with four equal subintervals. Is it an over- or underestimate?

   Show answer

   \(L_4 = f(0) + f(1) + f(2) + f(3) = 1 + 2 + 5 + 10 = 18\). \(f\) is increasing, so this underestimates.

2. Use a trapezoidal sum with the table to estimate \(\displaystyle\int_0^6 v(t)\,dt\): \(t = 0, 1, 3, 6\); \(v(t) = 10, 14, 20, 26\).

   Show answer

   \(\tfrac{10 + 14}{2}(1) + \tfrac{14 + 20}{2}(2) + \tfrac{20 + 26}{2}(3) = 12 + 34 + 69 = 115\).

3. Evaluate \(\displaystyle\int_1^3 (6x^2 - 2x + 1)\,dx\).

   Show answer

   \(\Big[2x^3 - x^2 + x\Big]_1^3 = (54 - 9 + 3) - (2 - 1 + 1) = 46\).

4. Evaluate \(\displaystyle\int_0^{\pi/3}\sec^2 x\,dx\).

   Show answer

   \(\Big[\tan x\Big]_0^{\pi/3} = \sqrt{3} - 0 = \sqrt{3}\).

5. Find \(\dfrac{d}{dx}\displaystyle\int_2^{x^3}\sqrt{1 + t^2}\,dt\).

   Show answer

   By the FTC with the chain rule: \(\sqrt{1 + x^6}\cdot 3x^2\).

6. Find \(\displaystyle\int x^2(x^3 + 5)^4\,dx\).

   Show answer

   \(u = x^3 + 5\), \(du = 3x^2\,dx\): \(\dfrac{1}{3}\cdot\dfrac{u^5}{5} = \dfrac{(x^3 + 5)^5}{15} + C\).

7. Evaluate \(\displaystyle\int_0^1 x\,e^{-x^2}\,dx\).

   Show answer

   \(u = -x^2\), \(du = -2x\,dx\). \(\Big[-\tfrac{1}{2}e^{-x^2}\Big]_0^1 = \dfrac{1 - e^{-1}}{2} \approx 0.316\).

8. Find \(\displaystyle\int\frac{x^2 + 2x + 3}{x + 1}\,dx\).

   Show answer

   Divide: \(x^2 + 2x + 3 = (x + 1)^2 + 2\), so the integrand is \(x + 1 + \dfrac{2}{x + 1}\). Result: \(\dfrac{x^2}{2} + x + 2\ln|x + 1| + C\).

9. Given \(\displaystyle\int_{-2}^{4} f(x)\,dx = 10\) and \(\displaystyle\int_{1}^{4} f(x)\,dx = 3\), find \(\displaystyle\int_{-2}^{1}\big(2f(x) + 3\big)\,dx\).

   Show answer

   \(\displaystyle\int_{-2}^{1} f = 10 - 3 = 7\). Then \(2(7) + 3(3) = 23\).

10. \(f(t) = 2\) on \([0, 3]\), then drops linearly from \((3, 2)\) to \((5, -2)\), then stays at \(-2\) on \([5, 7]\). Let \(g(x) = \displaystyle\int_0^x f(t)\,dt\). Find \(g(5)\), and the \(x\) in \([0, 7]\) where \(g\) is greatest.

    Show answer

    \(g(3) = 6\). From 3 to 4, \(f\) goes from 2 to 0 (area \(+1\)); from 4 to 5, from 0 to \(-2\) (area \(-1\)). So \(g(5) = 6\). \(g\) is greatest where \(f\) changes from positive to negative: \(x = 4\), with \(g(4) = 7\).

Lesson 7.1 · Unit 7 · CED topic 7.1

Modeling situations with differential equations

A differential equation is an equation involving a derivative. Most real laws are stated this way, not "here is the population" but "here is how fast the population is changing." The first skill is translating a sentence about rates into symbols.

Lesson 7.2 · Unit 7 · CED topic 7.2

Verifying solutions to differential equations

A solution to a differential equation is a function that makes the equation true. Checking a proposed solution needs no solving at all: differentiate it, substitute, and see whether both sides agree.

A general solution contains an arbitrary constant (\(y = Ce^{2x}\)); a particular solution has the constant pinned down by an initial condition (\(y = 3e^{2x}\)).

Lesson 7.3 · Unit 7 · CED topics 7.3–7.4

Sketching and reasoning with slope fields

A differential equation \(\dfrac{dy}{dx} = F(x, y)\) tells you the slope at every point in the plane. Draw a short segment with that slope at a grid of points and you have a slope field: a picture of every solution at once, before you've solved anything.

To sketch a particular solution through a given point, start at that point and follow the segments in both directions, staying tangent to the field. The curve never crosses the segments.

Lesson 7.4 · Unit 7 · CED topic 7.6

Finding general solutions using separation of variables

The one solving technique in AB Calculus: if the right side factors into a function of \(x\) times a function of \(y\), move everything involving \(y\) to the left with \(dy\), everything involving \(x\) to the right with \(dx\), and integrate both sides.

Lesson 7.5 · Unit 7 · CED topic 7.7

Finding particular solutions using initial conditions

An initial condition, one point the solution passes through, selects a single curve from the family of general solutions. Solve in general, substitute the point, and solve for \(C\). Then check the domain: the solution must be a continuous curve through the given point.

Lesson 7.6 · Unit 7 · CED topic 7.8

Exponential models with differential equations

The most important differential equation in the course is \(\dfrac{dy}{dt} = ky\): a quantity changing at a rate proportional to itself. Separation of variables solves it once and for all, and the solution describes populations, radioactive decay, interest, and cooling.

Unit 7 practice · 10 problems

Unit 7 practice: Differential Equations

Ten problems covering the whole unit. Work each one on paper before revealing the answer. A calculator is fine for the final arithmetic in problems 8 and 9.

1. Write a differential equation: the temperature \(T\) of a pie changes at a rate proportional to the difference between \(T\) and the 75° room.

   Show answer

   Translate phrase by phrase: the rate of change of \(T\) is \(\dfrac{dT}{dt}\); proportional to introduces a constant \(k\); the difference between \(T\) and the room is \(T - 75\). So \(\dfrac{dT}{dt} = k(T - 75)\), with \(k\) negative as the pie cools.

2. Verify that \(y = 2e^{3x} - 1\) is a solution of \(y' = 3y + 3\).

   Show answer

   \(y' = 6e^{3x}\). Right side: \(3(2e^{3x} - 1) + 3 = 6e^{3x}\). Equal. ✓

3. For \(\dfrac{dy}{dx} = x + y\), find the slopes at \((1, 2)\) and \((-2, 2)\), and the line along which the slope-field segments are horizontal.

   Show answer

   Slopes \(3\) and \(0\). Horizontal along \(y = -x\).

4. Find the general solution of \(\dfrac{dy}{dx} = \dfrac{4x^3}{y}\).

   Show answer

   \(\displaystyle\int y\,dy = \int 4x^3\,dx \Rightarrow \dfrac{y^2}{2} = x^4 + C\), or \(y^2 = 2x^4 + C\).

5. Solve \(\dfrac{dy}{dx} = 2y\) with \(y(0) = 3\).

   Show answer

   Separate the variables: \(\dfrac{dy}{y} = 2\,dx\). Integrate both sides: \(\ln|y| = 2x + C\), so \(y = Ae^{2x}\). Apply \(y(0) = 3\): \(A = 3\), giving \(y = 3e^{2x}\).

6. Solve \(\dfrac{dy}{dx} = \dfrac{x^2}{y^2}\) with \(y(0) = 2\).

   Show answer

   \(\dfrac{y^3}{3} = \dfrac{x^3}{3} + C\); \(C = \tfrac{8}{3}\). So \(y^3 = x^3 + 8\) and \(y = \left(x^3 + 8\right)^{1/3}\).

7. Solve \(\dfrac{dy}{dx} = (y - 3)\cos x\) with \(y(0) = 5\).

   Show answer

   \(\ln|y - 3| = \sin x + C\); at \((0, 5)\), \(C = \ln 2\). So \(y - 3 = 2e^{\sin x}\), i.e. \(y = 3 + 2e^{\sin x}\).

8. A radioactive sample decays from 80 g to 20 g in 6 hours. Find \(k\) in \(\dfrac{dy}{dt} = ky\), and the half-life.

   Show answer

   \(20 = 80e^{6k} \Rightarrow e^{6k} = \tfrac{1}{4} \Rightarrow k = -\dfrac{\ln 4}{6} = -\dfrac{\ln 2}{3} \approx -0.231\). Half-life \(= \dfrac{\ln 2}{|k|} = 3\) hours (80 → 40 → 20 in two half-lives).

9. A population satisfies \(\dfrac{dP}{dt} = 0.02P\) with \(P(0) = 1500\). When does it reach 3000?

   Show answer

   \(P = 1500e^{0.02t} = 3000 \Rightarrow t = \dfrac{\ln 2}{0.02} \approx 34.7\) years.

10. In a slope field, every segment in a given vertical column is parallel, and the segments along the \(y\)-axis are horizontal. Which equation matches?

    \(y' = y\) \(y' = x\) \(y' = xy\) \(y' = x - y\)

    If \(y' = y\), the slope depends on \(y\) alone, so the segments would be parallel along horizontal rows, not vertical columns, and along the \(y\)-axis the slopes would run from negative to positive instead of all being horizontal.

    Parallel segments within each vertical column mean the slope depends on \(x\) only, and horizontal segments where \(x = 0\) mean the slope is zero there. \(y' = x\) satisfies both: it is constant on any vertical line and equals zero exactly on the \(y\)-axis.

    \(y' = xy\) does give zero slope on the \(y\)-axis (where \(x = 0\)), which is why it tempts, but within a vertical column the slope still changes with \(y\), so the segments in a column would not be parallel.

    \(y' = x - y\) depends on both variables, so the segments in a column tilt differently as \(y\) changes, and on the \(y\)-axis the slope is \(-y\), which is horizontal only at the origin.

Lesson 8.1 · Unit 8 · CED topic 8.1

Finding the average value of a function

The average of a list of numbers is their sum divided by how many there are. The average of a function on an interval is the same idea with an integral in place of the sum: total accumulation divided by the length of the interval.

Lesson 8.2 · Unit 8 · CED topic 8.2

Connecting position, velocity, and acceleration using integrals

Unit 4 went from position to velocity to acceleration by differentiating. Integrals go the other way, and they make one distinction sharp: displacement versus total distance.

Lesson 8.3 · Unit 8 · CED topic 8.3

Using accumulation functions and definite integrals in applied contexts

The signature AB free-response problem: something flows in at one rate and out at another, and you're asked how much there is at a given time and when there's the most. The tools are net change and the first derivative test, applied to an integral.

Lesson 8.4 · Unit 8 · CED topic 8.4

Finding the area between curves expressed as functions of x

The area under one curve is an integral. The area between two curves is the integral of the gap between them: top minus bottom, over the \(x\)-interval where the region lives. Finding that interval usually means finding where the curves intersect.

Lesson 8.5 · Unit 8 · CED topics 8.5–8.6

Areas with functions of y, and regions with multiple intersections

Sometimes a region is awkward to slice vertically: a curve doubles back, or the "top" changes partway through. Slice horizontally instead, integrating right minus left with respect to \(y\). And when curves cross more than twice, split the integral wherever top and bottom trade places.

Lesson 8.6 · Unit 8 · CED topics 8.7–8.8

Volumes with cross sections

Slice a solid into thin slabs perpendicular to an axis. Each slab has volume (cross-sectional area) × (thickness). Add them up with an integral and you have the volume, no revolving required.

Lesson 8.7 · Unit 8 · CED topics 8.9–8.10

Volumes of revolution: the disc method

Revolve a region around a line and every cross section perpendicular to that line is a circle. The cross-section formula becomes \(A = \pi R^2\), where \(R\) is the distance from the axis of rotation to the curve, so the only new work is writing that radius correctly.

Lesson 8.8 · Unit 8 · CED topics 8.11–8.12

Volumes of revolution: the washer method

If the region doesn't touch the axis of rotation, revolving it leaves a hole. Each cross section is a washer, a disc with a smaller disc removed, and its area is \(\pi R^2 - \pi r^2\).

Unit 8 practice · 10 problems

Unit 8 practice: Applications of Integration

Ten problems covering the whole unit. Work each one on paper before revealing the answer. No calculator needed.

1. Find the average value of \(f(x) = 6x^2\) on \([0, 2]\).

   Show answer

   \(\dfrac{1}{2}\displaystyle\int_0^2 6x^2\,dx = \dfrac{1}{2}\Big[2x^3\Big]_0^2 = 8\).

2. \(v(t) = 3t^2 - 12\) on \([0, 3]\). Find the displacement and the total distance traveled.

   Show answer

   Displacement: \(\Big[t^3 - 12t\Big]_0^3 = -9\). \(v = 0\) at \(t = 2\): \(\displaystyle\int_0^2 v = -16\), \(\displaystyle\int_2^3 v = 7\). Total distance \(= 16 + 7 = 23\).

3. \(a(t) = 2t - 4\) and \(v(0) = 3\). Find \(v(t)\) and the displacement on \([0, 3]\).

   Show answer

   \(v(t) = t^2 - 4t + 3\). Displacement \(= \Big[\tfrac{t^3}{3} - 2t^2 + 3t\Big]_0^3 = 9 - 18 + 9 = 0\).

4. Water enters a tank at \(12 - t\) gal/min and leaves at 4 gal/min for \(0 \le t \le 12\); the tank holds 50 gal at \(t = 0\). When is the amount of water greatest, and how much is there then?

   Show answer

   Net rate \(8 - t\), positive then negative, zero at \(t = 8\). Amount \(= 50 + \displaystyle\int_0^8 (8 - t)\,dt = 50 + 32 = 82\) gallons at \(t = 8\).

5. Find the area between \(y = x^2 - 4\) and \(y = 2x - 1\).

   Show answer

   Intersections at \(x = -1, 3\); the line is on top. \(\displaystyle\int_{-1}^{3}\big(-x^2 + 2x + 3\big)\,dx = \Big[-\tfrac{x^3}{3} + x^2 + 3x\Big]_{-1}^{3} = 9 + \tfrac{5}{3} = \tfrac{32}{3}\).

6. Set up and evaluate an integral with respect to \(y\) for the area bounded by \(y = \sqrt{x}\), \(y = 0\), and \(x = 4\).

   Show answer

   Horizontal strips run from \(x = y^2\) to \(x = 4\) for \(0 \le y \le 2\): \(\displaystyle\int_0^2 (4 - y^2)\,dy = 8 - \tfrac{8}{3} = \tfrac{16}{3}\). (Check: \(\displaystyle\int_0^4\sqrt{x}\,dx = \tfrac{16}{3}\).)

7. The region bounded by \(y = x^2\), \(y = 0\), and \(x = 2\) is revolved about the \(x\)-axis. Find the volume.

   Show answer

   \(\pi\displaystyle\int_0^2 (x^2)^2\,dx = \pi\Big[\tfrac{x^5}{5}\Big]_0^2 = \dfrac{32\pi}{5}\).

8. The region between \(y = \sqrt{x}\) and \(y = \dfrac{x}{2}\) is revolved about the \(x\)-axis. Find the volume.

   Show answer

   They meet at \(x = 0\) and \(4\), with \(\sqrt{x}\) on top. \(\pi\displaystyle\int_0^4\left(x - \tfrac{x^2}{4}\right)dx = \pi\Big[\tfrac{x^2}{2} - \tfrac{x^3}{12}\Big]_0^4 = \pi\left(8 - \tfrac{16}{3}\right) = \dfrac{8\pi}{3}\).

9. The base of a solid is the region between \(y = x\) and \(y = x^2\). Cross sections perpendicular to the \(x\)-axis are equilateral triangles. Find the volume.

   Show answer

   Side \(s = x - x^2\), \(A = \tfrac{\sqrt{3}}{4}s^2\). \(V = \tfrac{\sqrt{3}}{4}\displaystyle\int_0^1 (x - x^2)^2\,dx = \tfrac{\sqrt{3}}{4}\left(\tfrac{1}{3} - \tfrac{1}{2} + \tfrac{1}{5}\right) = \tfrac{\sqrt{3}}{4}\cdot\tfrac{1}{30} = \dfrac{\sqrt{3}}{120}\).

10. The region between \(y = x^2\) and \(y = 1\) is revolved about the line \(y = 2\). Find the volume.

    Show answer

    Outer radius \(2 - x^2\), inner radius \(1\). \(V = \pi\displaystyle\int_{-1}^{1}\Big[(2 - x^2)^2 - 1\Big]dx = 2\pi\displaystyle\int_0^1\big(3 - 4x^2 + x^4\big)\,dx = 2\pi\cdot\tfrac{28}{15} = \dfrac{56\pi}{15}\).

Unit recap

Unit recap

Animated recap with on-screen narration. Turn on Voice to have it read aloud (uses your device's built-in voice). Pressing play counts as your one free video.