AP Calculus AB — Master Study Guide

1

PROBLEM 01

LIMITS

The 0/0 Limit — Don't Give Up!

Evaluate the limit. Most students immediately say "undefined." That's wrong — factor first.

lim (x² - 4) / (x - 2) as x \to 2

▸ EXAMPLE — Worked Approach

Factor the numerator: x² − 4 = (x+2)(x−2). Cancel (x−2) with the denominator. You're left with (x+2). Now plug in x = 2.

lim (x+2) = 2 + 2 = ? x\to2

⚠ COMMON TRAP
Plugging in x=2 first gives 0/0, which is an indeterminate form — NOT the final answer. Always try to factor/cancel before concluding.

▸ YOUR TURN — Practice

Find the limit:

lim (x² - 9) / (x - 3) as x \to 3

⚡ MEMORY POINT

When you see 0/0 → Try FACTOR → CANCEL → SUBSTITUTE
Magic word: F.C.S.

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PROBLEM 02

LIMITS

One-Sided Limits & Existence

Given the piecewise function below, does lim f(x) as x → 2 exist?

f(x) = { x + 1, if x < 2 { x² , if x \geq 2

▸ KEY IDEA

A two-sided limit exists only if both one-sided limits are equal.

lim⁻ f(x) = 2 + 1 = 3 lim⁺ f(x) = 2² = 4 3 \neq 4 \to Limit does NOT exist

⚠ COMMON TRAP
f(2) = 4 exists as a function value — but that doesn't mean the limit exists! Limit and function value are different things.

▸ YOUR TURN

g(x) = { 2x - 1, if x < 3 { x²- 4, if x \geq 3 Does lim g(x) exist as x \to 3?

⚡ MEMORY POINT

Limit EXISTS only when: LEFT = RIGHT
L⁻ = L⁺ → Limit exists | L⁻ ≠ L⁺ → DNE

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PROBLEM 03

LIMITS

Limits at Infinity — Horizontal Asymptotes

Evaluate:

lim (3x² + 5x) / (2x² - 1) as x \to \infty

▸ SHORTCUT METHOD

Compare the degree of numerator vs. denominator.

Same degree \to ratio of LEADING coefficients \to 3/2

⚠ COMMON TRAP
Students divide wrong or forget to compare degrees. Rules: same degree = coeff. ratio, top heavier = ∞, bottom heavier = 0.

▸ YOUR TURN

A) lim (5x³ + 2) / (x³ - 7) as x \to \infty B) lim (4x²) / (x³ + 1) as x \to \infty

⚡ MEMORY POINT

BOBO: Bottom degree bigger → answer is 0
TOBO: Top bigger → ∞ (no H.A.)
EATS DC: Equal degrees → divide the Coefficients

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PROBLEM 04

DERIVATIVES

Definition of the Derivative

Using the limit definition, find f'(x) for f(x) = x².

f'(x) = lim [f(x+h) - f(x)] / h as h\to0

▸ STEP-BY-STEP

= lim [(x+h)² - x²] / h = lim [x²+2xh+h² - x²] / h = lim [2xh + h²] / h = lim h(2x + h) / h = lim (2x + h) \to 2x

⚠ COMMON TRAP
Forgetting to expand (x+h)² correctly. (x+h)² = x² + 2xh + h² NOT x² + h²

▸ YOUR TURN

Use limit def. to find f'(x) for f(x) = 3x² - x

⚡ MEMORY POINT

Steps: PLUG → EXPAND → CANCEL h → SUBSTITUTE h=0

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PROBLEM 05

DERIVATIVES

Chain Rule — The Most Missed Rule

Differentiate: f(x) = (3x² + 1)⁵

▸ CHAIN RULE FORMULA

d/dx [g(h(x))] = g'(h(x)) \cdot h'(x) "Derivative of outside \times derivative of inside"

outer = u⁵, inner = 3x² + 1 f'(x) = 5(3x²+1)⁴ \cdot 6x = 30x(3x²+1)⁴

⚠ COMMON TRAP
Forgetting to multiply by the derivative of the inside. Writing 5(3x²+1)⁴ ONLY is the #1 chain rule mistake.

▸ YOUR TURN

A) y = sin(4x²) B) y = e^(x³ - 2x) C) y = \sqrt(5x + 3)

⚡ MEMORY POINT

OUTSIDE first (don't change inside) → then multiply by INSIDE'
Mantra: "Copy, Power, Inside Prime"

6

PROBLEM 06

DERIVATIVES

Implicit Differentiation

Find dy/dx given: x² + y² = 25

▸ KEY IDEA

When y is inside, its derivative picks up a dy/dx factor (chain rule with y as inside function).

Differentiate both sides w.r.t. x: 2x + 2y\cdot(dy/dx) = 0 dy/dx = -x / y

⚠ COMMON TRAP
Students differentiate y² as 2y, forgetting the dy/dx that must multiply it. Always tag dy/dx onto every y-term derivative.

▸ YOUR TURN

Find dy/dx: A) x³ + y³ = 6xy B) sin(x) + cos(y) = 1

⚡ MEMORY POINT

Rule: any time you differentiate y, slap a dy/dx next to it.
Then ISOLATE dy/dx to one side.

7

PROBLEM 07

DERIVATIVES

Related Rates

A spherical balloon is being inflated. At the moment its radius is 3 cm, the radius is increasing at 2 cm/s. How fast is the volume increasing?

V = (4/3)πr³

▸ STRATEGY

Differentiate both sides with respect to t: dV/dt = 4πr² \cdot dr/dt Plug in: r = 3, dr/dt = 2 dV/dt = 4π(9)(2) = ?

⚠ COMMON TRAP
Plugging in numbers BEFORE differentiating. Always differentiate first with variables, THEN substitute given values.

▸ YOUR TURN

A ladder 10 ft long leans against a wall. The bottom slides away at 1 ft/s. How fast is the top sliding down when the bottom is 6 ft away?

⚡ MEMORY POINT

Order: EQUATION → DIFFERENTIATE wrt t → PLUG IN
Never plug in before differentiating!

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PROBLEM 08

DERIVATIVES

Increasing, Decreasing & Critical Points

Find all critical points of f(x) = x³ − 6x² + 9x and determine where f is increasing or decreasing.

Step 1: f'(x) = 3x² - 12x + 9 Step 2: Set f'(x) = 0

3(x² - 4x + 3) = 0 3(x - 1)(x - 3) = 0 Critical pts: x = 1, x = 3

Test intervals: (−∞,1), (1,3), (3,∞)

⚠ COMMON TRAP
Calling every critical point a max or min without testing! Always do a sign chart or second derivative test.

▸ YOUR TURN

Find critical pts and increasing/decreasing intervals for: g(x) = 2x³ - 3x² - 12x + 1

⚡ MEMORY POINT

f' > 0 → INCREASING | f' < 0 → DECREASING
f' = 0 or undefined → CRITICAL POINT (not automatically a max/min!)

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PROBLEM 09

DERIVATIVES

Concavity & Inflection Points

For f(x) = x⁴ − 4x³, determine concavity and find all inflection points.

f'(x) = 4x³ - 12x² f''(x) = 12x² - 24x = 12x(x - 2)

f''(x) = 0 at x = 0 and x = 2 Test: concave up when f'' > 0 concave down when f'' < 0

⚠ COMMON TRAP
f''(c) = 0 does NOT automatically mean inflection point. You must verify that concavity actually CHANGES at c.

▸ YOUR TURN

Find inflection pts for h(x) = x⁵ - 5x⁴

⚡ MEMORY POINT

f'' > 0 → CONCAVE UP (smile 😊)
f'' < 0 → CONCAVE DOWN (frown 😔)
INFLECTION = where concavity changes sign

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PROBLEM 10

DERIVATIVES

Absolute Max & Min on a Closed Interval

Find the absolute maximum and minimum of f(x) = x³ − 3x on [−2, 2].

▸ CLOSED INTERVAL METHOD

1. Find f'(x) = 3x² - 3 = 0 \to x = \pm1 2. Evaluate f at critical pts AND endpoints: f(-2) = -8+6 = -2 f(-1) = -1+3 = 2 f( 1) = 1-3 = -2 f( 2) = 8-6 = 2 3. Largest = abs. MAX, Smallest = abs. MIN

⚠ COMMON TRAP
Forgetting to evaluate at the endpoints! Many students only check critical points and miss the true absolute extrema.

▸ YOUR TURN

Find abs. max & min of g(x) = x⁴ - 8x² on [-3, 3]

⚡ MEMORY POINT

C.E.P. = Critical pts + Endpoints + Pick largest/smallest
Always check ALL three locations!

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PROBLEM 11

INTEGRALS

Basic Antiderivatives — Power Rule

Evaluate the indefinite integral:

\int (4x³ - 6x + 5) dx

▸ POWER RULE FOR INTEGRALS

\int xⁿ dx = x^(n+1) / (n+1) + C ↑ Add 1 to exponent, divide by new exponent

= x⁴ - 3x² + 5x + C

⚠ COMMON TRAP
Forgetting + C on indefinite integrals! On AP exam, missing C loses points every time.

▸ YOUR TURN

A) \int (3x² + 2x - 1) dx B) \int (x⁴ - 5x² + x) dx

⚡ MEMORY POINT

Derivative: POWER DOWN (subtract 1)
Integral: POWER UP (add 1, divide) + +C !

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PROBLEM 12

INTEGRALS

U-Substitution

Evaluate:

\int 2x \cdot cos(x²) dx

▸ U-SUB STRATEGY

Look for something inside whose derivative is also outside.

Let u = x² \to du = 2x dx Rewrite: \int cos(u) du = sin(u) + C Back-sub: sin(x²) + C

⚠ COMMON TRAP
Forgetting to convert ALL x's to u (including dx → du). Also, forgetting to back-substitute at the end.

▸ YOUR TURN

A) \int 3x² \cdot e^(x³) dx B) \int sin(5x) dx C) \int x / (x²+1) dx

⚡ MEMORY POINT

Spot the INSIDE function → check if its DERIVATIVE is sitting outside → that's your u!
Steps: CHOOSE u → FIND du → SUBSTITUTE → BACK-SUB

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PROBLEM 13

INTEGRALS

Area Between Two Curves

Find the area enclosed between f(x) = x² and g(x) = x + 2.

1. Find intersections: x² = x + 2 \to x = -1, 2 2. Area = \int₋₁² [(x+2) - x²] dx 3. = [x²/2 + 2x - x³/3] from -1 to 2

⚠ COMMON TRAP
Subtracting in the wrong order! Area = ∫[TOP − BOTTOM]. Always check which curve is on top in the interval.

▸ YOUR TURN

Find the area between y = x² - 2 and y = -x² + 2

⚡ MEMORY POINT

Formula: ∫[TOP − BOTTOM] from a to b
Steps: INTERSECT → IDENTIFY TOP → INTEGRATE

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PROBLEM 14

FTC

Fundamental Theorem of Calculus — Part 1

If F(x) = ∫₀ˣ √(t² + 1) dt, find F'(x).

▸ FTC PART 1

d/dx [\intₐˣ f(t) dt] = f(x) \to Just plug in x for t! F'(x) = \sqrt(x² + 1)

▸ CHAIN RULE UPGRADE

What if the upper limit is a function of x?

d/dx [\int₀^(x²) \sqrt(t+1) dt] = \sqrt(x²+1) \cdot 2x ↑ chain rule!

⚠ COMMON TRAP
Forgetting the chain rule when the upper bound is NOT just "x" but a function like x², sin(x), etc.

▸ YOUR TURN

A) F(x) = \int₁ˣ (t³+1) dt Find F'(x) B) G(x) = \int₀^(x³) cos(t) dt Find G'(x)

⚡ MEMORY POINT

FTC1: PLUG x INTO integrand (forget the dt!)
If upper bound = function → multiply by CHAIN RULE

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PROBLEM 15

FTC

Fundamental Theorem of Calculus — Part 2

Evaluate the definite integral:

\int₁⁴ (2x + 3) dx

▸ FTC PART 2

= [x² + 3x] from 1 to 4 = (16 + 12) - (1 + 3) = 28 - 4 = ?

⚠ COMMON TRAP
Forgetting to subtract F(a) after computing F(b). Writing only F(b) is incomplete. Also: NO + C for definite integrals (they cancel).

▸ YOUR TURN

A) \int₀³ (x² - 2x + 1) dx B) \int₁⁴ (1/\sqrtx) dx

⚡ MEMORY POINT

FTC2: Antiderivative → F(b) − F(a)
"Top minus Bottom" → PLUG TOP, PLUG BOTTOM, SUBTRACT

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PROBLEM 16

INTEGRALS

Average Value of a Function

Find the average value of f(x) = x² on [1, 4].

f_avg = 1/(b-a) \cdot \intₐᵇ f(x) dx

= 1/(4-1) \cdot \int₁⁴ x² dx = (1/3) \cdot [x³/3] from 1 to 4 = (1/3) \cdot (64/3 - 1/3) = (1/3) \cdot (63/3) = 7

⚠ COMMON TRAP
Forgetting the 1/(b−a) multiplier in front. Students often just compute the integral and forget this "averaging" factor.

▸ YOUR TURN

Find avg value of g(x) = sin(x) on [0, π]

⚡ MEMORY POINT

AVG VALUE = 1/(b−a) × integral from a to b
Think: total area ÷ width of interval

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PROBLEM 17

DIFF. EQ.

Separable Differential Equations

Solve the differential equation with the given initial condition:

dy/dx = 2xy, y(0) = 3

▸ SEPARATE & INTEGRATE

dy/y = 2x dx \int dy/y = \int 2x dx ln|y| = x² + C y = Ae^(x²) Use y(0) = 3: 3 = Ae⁰ = A y = 3e^(x²)

⚠ COMMON TRAP
Forgetting to use the initial condition to find C (or A). And forgetting the absolute value in ln|y|.

▸ YOUR TURN

Solve: dy/dx = y/x, y(1) = 4

⚡ MEMORY POINT

Steps: SEPARATE (all y on left, all x on right)
→ INTEGRATE both sides → SOLVE for y → PLUG IN initial condition

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PROBLEM 18

DIFF. EQ.

Exponential Growth & Decay

A population grows at a rate proportional to its size. At t=0 the population is 500, and at t=3 it is 2000. Find the population at t=5.

Model: P(t) = P₀ \cdot e^(kt)

Step 1: 2000 = 500\cdote^(3k) e^(3k) = 4 k = ln(4)/3 Step 2: P(5) = 500\cdote^(5k) = 500\cdote^(5\cdotln4/3) = ?

⚠ COMMON TRAP
Using the wrong formula like P = P₀ + kt (linear). Proportional growth is ALWAYS exponential: P = P₀e^(kt).

▸ YOUR TURN

Radioactive substance has half-life 5 years. Starting with 200g, how much remains after 12 years?

⚡ MEMORY POINT

"proportional to itself" → dP/dt = kP → P = P₀e^(kt)
k > 0: growth | k < 0: decay

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PROBLEM 19

FTC

Riemann Sums → Definite Integral

Recognize the limit as a definite integral:

lim Σ (i/n)² \cdot (1/n) as n\to\infty, i=1 to n

▸ PATTERN RECOGNITION

Δx = 1/n, xᵢ = i/n This is: \int₀¹ x² dx = [x³/3]₀¹ = 1/3

The key: identify f(x) and the interval [a,b] from the sum structure.

⚠ COMMON TRAP
Confusing which part is f(xᵢ) and which is Δx. Remember: Δx = (b−a)/n, and xᵢ = a + i·Δx.

▸ YOUR TURN

Write as a definite integral: lim Σ \sqrt(1 + i/n) \cdot (1/n) n\to\infty, i=1 to n

⚡ MEMORY POINT

1/n = Δx → width, i/n = x → the sample point
Riemann sum IS the integral: lim Σ = ∫

20

PROBLEM 20

DERIVATIVES

Mean Value Theorem (MVT)

Verify MVT for f(x) = x³ − x on [0, 2], and find all values c that satisfy the conclusion.

MVT guarantees: f'(c) = [f(b)-f(a)] / (b-a)

f(0) = 0, f(2) = 6 Avg. rate = (6-0)/(2-0) = 3 f'(x) = 3x²-1 = 3 \to x² = 4/3 c = 2/\sqrt3 \approx 1.15 (in [0,2] ✓)

⚠ COMMON TRAP
Confusing MVT with Rolle's Theorem. Rolle's requires f(a) = f(b) and guarantees f'(c) = 0. MVT is more general.

▸ YOUR TURN

Find c satisfying MVT for f(x) = \sqrtx on [1, 9]

⚡ MEMORY POINT

MVT: If f is continuous on [a,b] and differentiable on (a,b),
then ∃c where f'(c) = SLOPE OF SECANT
Rolle's = MVT + f(a)=f(b) → f'(c)=0