AP Calculus AB · College Board CED Unit 5

Analytical Applications
of Differentiation

This is the biggest unit on the exam — 15–18% of your score. Everything here answers one question: where does the function reach its best value, and how do we prove it?

Press → to begin

38 SLIDES · 110 MIN · AP TOPIC 5

Unit 5 · 1 / 38

2 · Learning Objectives

By the end of this unit, you will be able to…

LO 1 State and apply the Extreme Value Theorem and the Mean Value Theorem, and know exactly what hypotheses each one needs.
LO 2 Find critical points and use them to determine intervals of increase and decrease.
LO 3 Classify local extrema with the First Derivative Test and absolute extrema with the Candidates Test.
LO 4 Determine concavity and locate inflection points using f''.
LO 5 Sketch f, f′, and f″ from a graph of any one of them — in both directions.
LO 6 Solve optimization problems with a complete, graded-FRQ-ready justification.

📊

Weight: Unit 5 is 15–18% of the multiple-choice section and appears on essentially every free-response question in some form. It is the single highest-leverage unit in AP Calculus AB.

Unit 5 · 2 / 38

3 · AP CED Mapping

Unit 5 → AP CED Topic 5

Twelve topics in one unit. Here is exactly where each one lives in this deck.

Topic	CED Skill ID	Slides
5.1 Using the Mean Value Theorem	FUN-1.B / FUN-4.A	7–10
5.2 Extreme Value Theorem, global vs local extrema, critical points	FUN-1.C / FUN-4.A	5–6, 14–15
5.3 Determining intervals of increase & decrease	FUN-4.A	11
5.4 Using the First Derivative Test	FUN-4.A	12–13
5.5 Using the Candidates Test	FUN-4.A	14–15
5.6 Determining concavity	FUN-4.B	16–17
5.7 Using the Second Derivative Test	FUN-4.B	18–19
5.8 Sketching graphs of f and f′	FUN-4.C	20–21
5.9 Connecting f, f′, and f″	FUN-4.C	22–23
5.10–5.11 Introduction to & solving optimization	FUN-4.B	25–29
5.12 Exploring behaviors of implicit relations	FUN-4.D	30–31

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4 · Key Terms

Eight terms you must be able to define on demand

Critical point

An interior point x = c in the domain of f where f′(c) = 0 or f′(c) does not exist.

Local (relative) extremum

f(c) is a local max/min if it is the largest/smallest value of f near c.

Absolute (global) extremum

f(c) is the largest/smallest value of f on its entire domain or on a given interval.

Concave up / down

f′ increasing (f″ > 0) / f′ decreasing (f″ < 0). Think "holds water" vs "spills water."

Inflection point

A point where f is continuous and concavity changes sign.

Candidates Test

Compare f at all critical points and both endpoints; the largest is the absolute max.

Secant line

Line through two points on the curve. Its slope = average rate of change over [a, b].

Optimization

Finding the input that maximizes or minimizes a quantity, subject to a constraint.

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5 · Topic 5.2 · Extreme Value Theorem · FUN-1.C

The Extreme Value Theorem (EVT)

Theorem · FUN-1.C.1

If f is continuous on a closed interval [a, b], then f attains both an absolute maximum and an absolute minimum on [a, b]. Each occurs at a critical point or at an endpoint.

✅ Both hypotheses required

Continuous and closed interval. Drop either one and the theorem gives you nothing.

❌ What breaks it

f(x) = 1/x on [−1, 1] is not continuous at 0 → no max. f(x) = x on (0, 1) is not closed → no max or min.

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6 · Topic 5.2 · Critical Points · FUN-4.A

Critical points: the only interior candidates

Definition · FUN-4.A.4

x = c is a critical point of f ⟺ f′(c) = 0 or f′(c) DNE, with c in the domain of f

Absolute extrema live at critical points or endpoints. Nothing else matters.

Type 1 · f′(c) = 0

Horizontal tangent. Smooth peak, valley, or a terrace (like x³ at 0).

Type 2 · f′(c) DNE

Corner (|x|), cusp, or vertical tangent (x^(1/3)). Still a critical point because c is in the domain.

⚠️

Trap: x = c where f(c) is undefined is NOT a critical point — it's not in the domain. It may be a vertical asymptote instead. Always check the domain first.

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7 · Topic 5.1 · Rolle's Theorem

Rolle's Theorem — the warm-up for MVT

Theorem

If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists at least one c in (a, b) with f′(c) = 0.

Geometrically: if the two endpoints are at the same height, the graph must turn around somewhere in between.

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8 · Topic 5.1 · Mean Value Theorem · FUN-1.B

The Mean Value Theorem (MVT)

Theorem · FUN-1.B.1

f′(c) = f(b) − f(a) ⁄ b − a

If f is continuous on [a, b] and differentiable on (a, b), then some c in (a, b) has instantaneous rate = average rate over [a, b].

🎯

Exam phrasing: "the tangent line is parallel to the secant line." When you see that sentence, they want MVT.

😐

Meme break

The Mean Value Theorem proves that at some instant, you were exactly average. Comforting and terrifying — just like a report card.

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9 · Mean Value Theorem · Worked

Three MVT problems

Worked A · find c

Q1. f(x) = x³ − 3x on [−2, 2]. Find all c satisfying MVT.

1 f is a polynomial → continuous & differentiable everywhere. Hypotheses ✓

2 Average rate = [f(2) − f(−2)] / [2 − (−2)] = [2 − (−2)] / 4 = 4/4 = 1.

3 f′(x) = 3x² − 3 = 1 → 3x² = 4 → x = ±2/√3 ≈ ±1.155.

✓ Both lie in (−2, 2). Answer: c = ±2/√3.

Worked B · check hypotheses

Q2. Does MVT apply to f(x) = |x| on [−1, 3]?

→ Continuous ✓ but NOT differentiable at x = 0, which is inside (−1, 3). MVT does not apply.

Worked C · justify existence

Q3. A car travels 120 miles in 2 hours. Prove its speed was exactly 60 mph at some instant.

→ s(t) is continuous & differentiable; avg rate = 120/2 = 60 mph. By MVT, ∃ c with s′(c) = 60 mph. This is exactly why speed cameras work.

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10 · Checkpoint · MVT & EVT

Quick check: MVT & EVT

Let f be differentiable on (0, 8) and continuous on [0, 8], with f(0) = 3 and f(8) = 19. Which of the following must be true?

Af′(c) = 2 for some c in (0, 8)

Bf′(c) = 2 for at least one c in (0, 8)

Cf has an absolute maximum on [0, 8] at a critical point

Df′(x) > 0 for all x in (0, 8)

✅ Answer: B. MVT gives f′(c) = (19 − 3)/(8 − 0) = 16/8 = 2 for at least one c — the word "at least" is what makes B safe. A says "= 2 for some c" which sounds similar but is a weaker form AP avoids; B is the College Board's standard wording. C is wrong — the absolute max could be at an endpoint. D is wrong — MVT guarantees the average is hit once, not that the function is always increasing.

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11 · Topic 5.3 · Increasing & Decreasing · FUN-4.A

The sign of f′ tells you which way f goes

Test · FUN-4.A.5 / FUN-4.A.6

f′(x) > 0 on (a, b) ⟹ f increasing | f′(x) < 0 on (a, b) ⟹ f decreasing

Find critical points → they split the number line → test one point per interval.

(−∞, −1)

−1

(−1, 2)

(2, ∞)

f′(x)

+ ↗

− ↘

+ ↗

f(x)

increasing

local max

decreasing

increasing

Write intervals with parentheses at critical points — f is increasing on the open interval, not "at" the point.
On the AP exam, "increasing on (−∞, −1) ∪ (2, ∞)" — never join with a ∪ across a decreasing gap unless they ask for it.

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12 · Topic 5.4 · First Derivative Test · FUN-4.A

First Derivative Test — classify by sign change

Test · FUN-4.A.7

At a critical point c with f continuous:
• f′ changes + → − at c ⟹ local maximum
• f′ changes − → + at c ⟹ local minimum
• f′ does not change sign ⟹ neither (a terrace)

🏔️

Meme break

Optimization problems ask "what is the maximum area for a given perimeter?" — finally, a math problem that respects fence budgets. (Farmers have been doing AP Calc for centuries; they just called it "not wasting fence.")

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13 · First Derivative Test · Worked

Full worked example with sign chart

Worked

Q. f(x) = x³ − 6x² + 9x + 1. Find intervals of increase/decrease and all local extrema.

1 f′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3).

2 Critical points: x = 1 and x = 3.

3 Sign chart: f′ > 0 on (−∞, 1); f′ < 0 on (1, 3); f′ > 0 on (3, ∞).

4 x = 1: f′ changes + → − ⟹ local max. f(1) = 1 − 6 + 9 + 1 = 5.

5 x = 3: f′ changes − → + ⟹ local min. f(3) = 27 − 54 + 27 + 1 = 1.

✓ Increasing on (−∞, 1) ∪ (3, ∞); decreasing on (1, 3). Local max (1, 5); local min (3, 1).

(−∞, 1)

(1, 3)

(3, ∞)

f′(x)

−

f(x)

↗ inc

MAX 5

↘ dec

↗ inc

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14 · Topic 5.5 · Candidates Test · FUN-4.A

Candidates Test — absolute extrema on a closed interval

Procedure · FUN-4.A.8

To find the absolute max & min of continuous f on [a, b]:
1. Find all critical points c in (a, b).
2. Evaluate f at every critical point and at both endpoints a, b.
3. The largest value is the absolute max; the smallest is the absolute min.

Why endpoints matter

f(x) = x on [0, 3] has no critical points at all, yet it has both an absolute max (3) and min (0) — at the endpoints.

On an open / infinite domain

There are no endpoints to test. Instead compare critical-point values and the limits as x → ±∞ to decide whether the max/min exists at all.

📝

FRQ wording: "Find the absolute maximum value of f on the closed interval [a, b]. Justify." You must list every candidate and its f-value, then state the winner. Skipping the list costs you the justification point.

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15 · Candidates Test · Worked

Worked: absolute extrema on [−1, 4]

Worked

Q. f(x) = 2x³ − 9x² + 12x + 1 on [−1, 4]. Find the absolute max and min.

1 f′(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2).

2 Critical points: x = 1, x = 2 — both inside [−1, 4]. ✓

3 Evaluate all four candidates:

Candidate x	Type	f(x)
x = −1	endpoint	f(−1) = −2 − 9 − 12 + 1 = −22	← absolute min
x = 1	critical	f(1) = 2 − 9 + 12 + 1 = 6	local max
x = 2	critical	f(2) = 16 − 36 + 24 + 1 = 5	local min
x = 4	endpoint	f(4) = 128 − 144 + 48 + 1 = 33	← absolute max

✅

Absolute min = −22 at x = −1. Absolute max = 33 at x = 4. Note both winners are endpoints — this is common and is exactly why the Candidates Test insists you check them.

Original comic · ap-study.com

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16 · Topic 5.6 · Concavity · FUN-4.B

Concavity — the sign of f″

Definition · FUN-4.B.3

f″(x) > 0 ⟹ concave up (f′ increasing) | f″(x) < 0 ⟹ concave down (f′ decreasing)

Concave up holds water. Concave down spills water.

🚿

Meme break

Concave up holds water and smiles. Concave down spills water and frowns. Calculus is just emotional plumbing.

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17 · Topic 5.6 · Inflection Points · FUN-4.B

Inflection points: continuity + sign change

Definition · FUN-4.B.4

An inflection point is a point on the graph of f where f is continuous and the concavity changes sign (f″ goes + → − or − → +).

✅ It IS an inflection point

f(x) = x³ at x = 0: f″(x) = 6x changes − → +, and f is continuous. Even though f′(0) = 0, that's a separate fact.

❌ It is NOT

f(x) = 1/x at x = 0: f″ = 2/x³ changes sign, but f is discontinuous there. Not an inflection point — it's a vertical asymptote.

Classic trap

"f″(c) = 0 therefore c is an inflection point." False. f″(c) = 0 is necessary but not sufficient. f(x) = x⁴ has f″(0) = 0 but is concave up on both sides — no sign change, no inflection point.

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18 · Topic 5.7 · Second Derivative Test · FUN-4.B

Second Derivative Test — the fast classifier

Test · FUN-4.B.5

f′(c) = 0 and f″(c) < 0 ⟹ local MAX | f′(c) = 0 and f″(c) > 0 ⟹ local MIN

If f″(c) = 0 the test is inconclusive — fall back to the First Derivative Test.

f′(c)

f″(c)

+ ∪

− ∩

DNE

conclusion

local MIN

local MAX

inconclusive

⚡

Which test should I use? If f″ is easy to compute, use the Second Derivative Test (one evaluation, done). If f″ is messy or zero, use the First Derivative Test — it always works.

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19 · Second Derivative Test · Failure Cases

Same f″(0) = 0, three different answers

This is why "f″(c) = 0 ⟹ nothing" is more accurate than "f″(c) = 0 ⟹ inflection."

Function	f′(0)	f″(0)	Behaviour at x = 0
f(x) = x⁴	0	0	local min — concave up both sides
f(x) = −x⁴	0	0	local max — concave down both sides
f(x) = x³	0	0	neither — inflection point, increasing throughout

Worked

Q. f(x) = x⁴ − 4x³. Classify all critical points.

1 f′(x) = 4x³ − 12x² = 4x²(x − 3). Critical points: x = 0, x = 3.

2 f″(x) = 12x² − 24x = 12x(x − 2). f″(3) = 12(3)(1) = 36 > 0 ⟹ local min at x = 3.

3 f″(0) = 0 ⟹ inconclusive. Use FDT: f′ = 4x²(x − 3) is negative on both sides of 0 (x² ≥ 0, x − 3 < 0).

✓ No sign change at 0 ⟹ neither max nor min. (It's an inflection point with a horizontal tangent.)

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20 · Topic 5.8 · Sketching f from f′ · FUN-4.C

Given the graph of f′, reconstruct f

f′ > 0 (above the x-axis) ⟹ f increasing. f′ < 0 ⟹ f decreasing.
f′ crosses the x-axis ⟹ f has a local extremum there (+ → − max; − → + min).
f′ increasing (going up) ⟹ f concave up. f′ decreasing ⟹ f concave down.
f′ has a local extremum ⟹ f has an inflection point at that same x.

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21 · Topic 5.8 · Sketching f′ from f · FUN-4.C

The reverse direction: read slopes off f

Read it in 3 passes

1️⃣ Mark where f is flat → those x-values are the zeros of f′. 2️⃣ Note where f rises (f′ > 0) vs falls (f′ < 0). 3️⃣ Note where f is steepest → f′ is farthest from the axis.

Where f has a corner

f′ has a jump discontinuity or DNE there. Draw an open circle, not a point.

💡

Mental shortcut: f′ is the "slope meter." Imagine walking along f and reading the incline at every step — that reading, plotted against x, is f′.

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22 · Topic 5.9 · f · f′ · f″ · FUN-4.C

The master table: f, f′, f″

If…	then f is…	f′ is…	f″ is…
f increasing	↗	positive	could be either
f decreasing	↘	negative	could be either
f concave up	∪	increasing	positive
f concave down	∩	decreasing	negative
f has a local max/min	peak / valley	zero or DNE	negative / positive (if f″ ≠ 0)
f has an inflection point	changes concavity	has a local extremum	zero or DNE, sign changes

🔑

The one fact students forget: f increasing and f concave up are independent. f(x) = e^x is both. f(x) = √x is increasing but concave down. f(x) = x² is decreasing and concave up on (−∞, 0). Never conflate the sign of f′ with the sign of f″.

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23 · Justification Language · AP Grader Gold

The exact sentences that earn FRQ points

AP graders do not read your mind. They read for these specific claim-and-reason pairs.

Claim	Required reason (must include the derivative!)
f has a local max at x = c	"because f′ changes from positive to negative at x = c"
f is increasing on (a, b)	"because f′(x) > 0 for all x in (a, b)"
f is concave up on (a, b)	"because f′ is increasing on (a, b)" (or f″ > 0)
f has an inflection point at x = c	"because f″ changes sign at x = c and f is continuous there"
f has an absolute max at x = c on [a, b]	"because f(c) is the largest among all critical points and endpoints"
There exists c with f′(c) = m	"by the Mean Value Theorem, since f is continuous on [a, b] and differentiable on (a, b)"

What loses the point

"f has a maximum at x = 2 because the graph goes up then down." No credit. You must name f′ and state its sign change. Substitute: "f′(x) changes from positive to negative at x = 2."

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24 · Checkpoint · Reading derivative graphs

AP-style multiple choice

The graph of f′ is positive on (−3, 1), negative on (1, 4), and positive on (4, 6). Also f′ is decreasing on (−3, 2) and increasing on (2, 6). How many inflection points does f have on (−3, 6)?

AZero

BOne

CTwo

DThree

✅ Answer: B. Inflection points of f occur where f′ changes from increasing to decreasing or vice versa — i.e. where f′ has a local extremum. f′ switches from decreasing to increasing exactly once, at x = 2. So f has exactly one inflection point. The sign changes of f′ (at x = 1 and x = 4) give local extrema of f, not inflection points — a very common mix-up.

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25 · Topic 5.10 · Introduction to Optimization

Optimization = maximize or minimize under a constraint

The structure of every optimization problem

You have a quantity to optimize (the objective function) and a constraint that limits your choices. The constraint lets you eliminate one variable so the objective becomes a function of a single variable.

Objective function

The thing you're maximizing or minimizing: area, volume, cost, revenue, distance, time.

Constraint

The fixed resource: fixed perimeter, fixed amount of material, fixed surface area, the curve the point must lie on.

🎯

What the AP grader checks: (1) correct primary equation with variables identified, (2) correct domain, (3) derivative set to zero, (4) the value of the answer with units, (5) justification that it's a max/min. Miss #5 and you lose a point even with the right number.

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26 · Topics 5.10–5.11 · The Recipe

Six steps that solve every optimization problem

Step 1 — Draw & label. Sketch the situation. Label every variable with its units. Never skip this.
Step 2 — Write the objective. "Maximize A = …" or "Minimize C = …". Name the function.
Step 3 — Write the constraint and substitute. Use it to eliminate one variable so the objective is a function of one variable.
Step 4 — State the domain. Usually 0 < x < something. This matters for the Candidates Test later.
Step 5 — Differentiate and find critical points. Set the derivative to zero (or find where it DNE), keep only the ones in the domain.
Step 6 — Justify and answer in context. Use the Candidates Test or Second Derivative Test, then write the answer as a sentence with units.

✍️

Sentence template for step 6: "Because A′(x) changes from positive to negative at x = …, A has a maximum there. The maximum area is … square feet when the dimensions are … by …."

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27 · Optimization · Worked Example 1

Classic: the open-top box

Worked

Q. A 12 in × 12 in sheet of cardboard has equal squares of side x cut from each corner, then the sides are folded up to make an open box. What x maximizes the volume?

1 Base becomes (12 − 2x) by (12 − 2x); height is x.

2 Objective: V(x) = x(12 − 2x)² · Domain: 0 < x < 6.

3 V(x) = x(144 − 48x + 4x²) = 144x − 48x² + 4x³.

4 V′(x) = 144 − 96x + 12x² = 12(x² − 8x + 12) = 12(x − 2)(x − 6).

5 Critical points: x = 2 (in domain) and x = 6 (not in the open domain — it gives V = 0).

6 Candidates Test: V(2) = 2(8)² = 128 in³; V → 0 as x → 0⁺ or x → 6⁻.

✓ Cut squares of side 2 inches. Max volume 128 in³, box is 8 × 8 × 2.

📦

Meme break

Nobody in recorded history has needed a maximum-volume lidless cardboard box. Calculus: solving imaginary crises since 1684 — and the AP exam still bills it as "real-world."

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28 · Optimization · Worked Example 2

Two more: fencing and cylinders

A · Fencing

Q1. 600 m of fencing encloses a rectangular field against a straight river (no fence needed along the river). Maximize the area.

→ Let x = width (perpendicular to river). Then length along river = 600 − 2x.

→ A(x) = x(600 − 2x) = 600x − 2x², domain 0 < x < 300.

→ A′(x) = 600 − 4x = 0 → x = 150 m; river side = 300 m. Max A = 45,000 m².

✓ A″(x) = −4 < 0 confirms a maximum.

B · Cylinder

Q2. A closed cylindrical can must hold 500 cm³. Minimize the surface area.

→ Constraint: πr²h = 500 → h = 500/(πr²).

→ S(r) = 2πr² + 2πrh = 2πr² + 1000/r.

→ S′(r) = 4πr − 1000/r² = 0 → 4πr³ = 1000 → r = ∛(250/π) ≈ 4.30 cm.

✓ h = 500/(π·4.30²) ≈ 8.60 cm. Notice h = 2r — a can that's as tall as it is wide is the most material-efficient. That's why real soup cans look like that.

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29 · Checkpoint · Optimization

AP-style multiple choice: optimization

A rectangle has its base on the x-axis and its two upper corners on the parabola y = 12 − x². What is the maximum possible area of such a rectangle?

A16

B32

C24

D48

✅ Answer: B (32). By symmetry let the upper-right corner be (x, 12 − x²) with x > 0. Width = 2x, height = 12 − x². A(x) = 2x(12 − x²) = 24x − 2x³. A′(x) = 24 − 6x² = 0 → x² = 4 → x = 2. A(2) = 2(2)(12 − 4) = 4 · 8 = 32. A″(x) = −12x < 0 at x = 2, so it's a maximum.

Unit 5 · 29 / 38

30 · Topic 5.12 · Behaviors of Implicit Relations · FUN-4.D

Analyzing curves that aren't functions

An implicit relation like x² + y² = 25 can fail the vertical line test — but you can still find where it's increasing, decreasing, concave up, and where it has vertical or horizontal tangents.

Horizontal tangent

dy/dx = 0. On an implicit curve: set the numerator of dy/dx equal to zero (and denominator ≠ 0).

Vertical tangent

dy/dx is undefined because the denominator is 0 while the numerator ≠ 0.

Increasing / decreasing

Same rules as always: dy/dx > 0 ⟹ increasing as x increases; dy/dx < 0 ⟹ decreasing.

Concavity

Compute d²y/dx² by differentiating dy/dx implicitly again, substituting dy/dx where it appears.

🧭

Strategy: with implicit relations, work branch by branch. A circle has an upper branch (y > 0) and a lower branch (y < 0), and their increasing/decreasing behavior is opposite.

Unit 5 · 30 / 38

31 · Topic 5.12 · Worked

Worked: x² + y² = 25, full analysis

Worked

Q. For the circle x² + y² = 25, find dy/dx, the points with horizontal tangents, the points with vertical tangents, and the concavity of the upper branch.

1 Differentiate: 2x + 2y·y′ = 0 → y′ = −x/y.

2 Horizontal tangents: y′ = 0 → x = 0 → points (0, 5) and (0, −5).

3 Vertical tangents: denominator y = 0 → points (5, 0) and (−5, 0).

4 Second derivative: y″ = −(y − x·y′)/y² = −(y − x(−x/y))/y² = −(y² + x²)/y³ = −25/y³.

5 Upper branch (y > 0): y″ = −25/y³ < 0 ⟹ concave down everywhere. ✓ (an arch)

6 Lower branch (y < 0): y″ > 0 ⟹ concave up. Upper branch is decreasing on (−5, 0) and increasing on (0, 5)? Check: y′ = −x/y with y > 0 → y′ > 0 when x < 0. So the upper branch increases on (−5, 0) and decreases on (0, 5), peaking at (0, 5).

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32 · Common Mistakes

Five mistakes that cost the most points in Unit 5

Confusing f′ sign with f″ sign

"Increasing" is about f′. "Concave up" is about f″. A function can be decreasing and concave up at the same time (like x² on (−∞, 0)).

Forgetting endpoints in the Candidates Test

On [a, b], the absolute max is very often at an endpoint. If you only check critical points you will get it wrong — and lose the answer point entirely.

Saying "f″(c) = 0 ⟹ inflection point"

You must confirm a sign change of f″ and continuity of f. x⁴ at 0 is the standard counterexample.

Claiming a max with no justification

"x = 3 maximizes the volume" earns zero justification credit. Write: "V′ changes from + to − at x = 3, so V has a maximum there."

Using MVT without checking hypotheses

MVT requires continuity on the closed interval and differentiability on the open interval. If either fails, MVT gives you nothing — say so explicitly.

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33 · AP Exam · Multiple Choice Practice

Two released-style MCQs

MCQ 1 · no calculator

1. Let f(x) = x³ − 3x² − 9x + 4. On which interval is f decreasing?

→ f′(x) = 3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1). f′ < 0 when −1 < x < 3. Answer: (−1, 3).

MCQ 2 · graph-based

2. The graph of the derivative g′ of a function g is shown above the x-axis on (0, 2), below on (2, 5), and above on (5, 7). If g(0) = 1, which is true?

→ g increases on (0, 2), decreases on (2, 5), increases on (5, 7). So g has a local max at x = 2 and a local min at x = 5. Values of g(7) vs g(0) cannot be determined without areas under g′ — that's Unit 6 territory.

⏱️

Timing: Unit 5 MCQs should take you 60–90 seconds each. If you're past 2 minutes, mark it and move on — the last 5 questions are usually worth the same as the first 5.

Unit 5 · 33 / 38

34 · AP Exam · FRQ Walkthrough

FRQ walkthrough: optimization with justification

FRQ · typical 9-point rubric

Prompt. A rectangular poster is to have an area of 180 in² with 1-inch margins at the top and bottom and 2-inch margins on each side. Find the dimensions of the poster that maximize the printed area.

1 pt Variables: let x = poster width, y = poster height. Constraint: xy = 180.

1 pt Objective: printed area P = (x − 4)(y − 2).

1 pt Substitute: y = 180/x → P(x) = (x − 4)(180/x − 2) = 180 − 2x − 720/x + 8 = 188 − 2x − 720/x.

1 pt Domain: x > 4.

2 pts Derivative: P′(x) = −2 + 720/x². Set = 0: 720/x² = 2 → x² = 360 → x = 6√10 ≈ 18.97 in.

1 pt Height: y = 180/x = 180/(6√10) = 30/√10 = 3√10 ≈ 9.49 in.

2 pts Justification: P′(x) > 0 for 4 < x < 6√10 and P′(x) < 0 for x > 6√10, so P has a maximum at x = 6√10. Dimensions: 6√10 in × 3√10 in.

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Note the exact form: AP accepts 6√10 or 18.974 but you should give 3 decimal places if you round. Never round intermediate values.

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35 · Real-World Applications

Optimization is everywhere once you see it

📦 Packaging & logistics

Amazon sizes boxes to minimize cardboard per unit volume. The math is exactly the cylinder problem on slide 28 — that's why shipping boxes cluster near a few standard aspect ratios.

💊 Drug dosage

Pharmacokinetics models blood concentration C(t). The peak (maximum) concentration and the time it occurs are found by setting C′(t) = 0. Too high = toxic; too low = ineffective.

💰 Marginal analysis in econ

Profit is maximized when marginal revenue = marginal cost, i.e. R′(q) = C′(q). That's setting P′(q) = 0 with P = R − C — the same first-derivative condition.

🚗 Traffic enforcement

Average-speed cameras are pure MVT: if you covered 120 miles in 2 hours, at some instant your speedometer read exactly 60 mph. The theorem is the legal argument.

Unit 5 · 35 / 38

36 · Video Resources

Three videos worth your time

Concept · 18 min

Mean Value Theorem — Complete Explanation

Professor Leonard · Full derivation + several worked examples with graphs

Worked examples · 21 min

Optimization Problems in Calculus

The Organic Chemistry Tutor · 8 classic problem types step by step

Intuition · 16 min

Derivatives Through Geometry | Essence of Calculus, Ch. 3

3Blue1Brown · Why the derivative rules look the way they do

Also search AP Classroom for the Unit 5 progress check videos — College Board's own videos match the exact wording used on the exam.

Unit 5 · 36 / 38

37 · Practice Problems

Six problems — try before you peek

Find the absolute max and min of f(x) = x⁴ − 8x² + 3 on [−3, 1].
Find all c satisfying MVT for f(x) = x² − 4x + 1 on [0, 5].
For f(x) = x⁴ − 4x³ + 10, find all intervals of concavity and all inflection points.
Find two positive numbers whose sum is 20 and whose product is as large as possible.
A 10 ft ladder leans against a wall. Find the maximum area of the triangle formed by the ladder, wall, and floor.
Given f′(x) = (x − 1)(x + 2)(x − 4), list the intervals where f increases, and classify each critical point.

✅

Answers: 1) max 6 at x = −3, min −13 at x = −2. 2) c = 5/2. 3) concave up on (−∞, 0) ∪ (2, ∞), down on (0, 2); inflection at x = 0 and x = 2. 4) 10 and 10. 5) 25 ft² (isosceles right triangle, 45°). 6) increase on (−2, 1) ∪ (4, ∞); local min at x = −2 and x = 4, local max at x = 1.

Unit 5 · 37 / 38

38 · Coverage Checklist

Unit 5 · complete topic coverage

Every CED learning objective for Unit 5, mapped to where it appears in this deck.

FUN-1.B.1 Mean Value Theorem — statement, hypotheses, geometric meaning (slides 8–9)
FUN-1.B.2 Rolle's Theorem as the special case of MVT (slide 7)
FUN-1.C.1 Extreme Value Theorem on a closed interval (slide 5)
FUN-4.A.4 Definition of a critical point (f′ = 0 or DNE) (slide 6)
FUN-4.A.5 / .6 Intervals of increase and decrease from the sign of f′ (slides 11, 13)
FUN-4.A.7 First Derivative Test for local extrema (slides 12–13)
FUN-4.A.8 Candidates Test for absolute extrema (slides 14–15)
FUN-4.B.3 Concavity from the sign of f″ (slide 16)
FUN-4.B.4 Inflection points require continuity + sign change (slide 17)
FUN-4.B.5 Second Derivative Test and its failure cases (slides 18–19)
FUN-4.C.1 / .2 Sketching f from f′ and f′ from f (slides 20–21)
FUN-4.C.3 Connecting the graphs of f, f′, and f″ (slide 22)
FUN-4.B.6 Optimization: set up, domain, solve, justify (slides 25–28)
FUN-4.D.1 / .2 Behaviors of implicitly defined relations (slides 30–31)
Justification language for every FRQ claim in this unit (slide 23)
AP exam formats — MCQ timing, FRQ rubric, 6 practice problems (slides 32–34, 37)

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Unit 5 complete. Next up: Unit 6 — Integration and Accumulation of Change. The second half of the course flips every question in this unit around: instead of "how fast is it changing," you'll ask "how much has accumulated?"

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Unit 5 1 / 38 Analytical Applications of Differentiation

Analytical Applicationsof Differentiation

By the end of this unit, you will be able to…

Unit 5 → AP CED Topic 5

Eight terms you must be able to define on demand

Critical point

Local (relative) extremum

Absolute (global) extremum

Concave up / down

Inflection point

Candidates Test

Secant line

Optimization

The Extreme Value Theorem (EVT)

✅ Both hypotheses required

❌ What breaks it

Critical points: the only interior candidates

Type 1 · f′(c) = 0

Type 2 · f′(c) DNE

Rolle's Theorem — the warm-up for MVT

The Mean Value Theorem (MVT)

Three MVT problems

Quick check: MVT & EVT

The sign of f′ tells you which way f goes

First Derivative Test — classify by sign change

Full worked example with sign chart

Candidates Test — absolute extrema on a closed interval

Why endpoints matter

On an open / infinite domain

Worked: absolute extrema on [−1, 4]

Concavity — the sign of f″

Inflection points: continuity + sign change

✅ It IS an inflection point

❌ It is NOT

Second Derivative Test — the fast classifier

Same f″(0) = 0, three different answers

Given the graph of f′, reconstruct f

The reverse direction: read slopes off f

Read it in 3 passes

Where f has a corner

The master table: f, f′, f″

The exact sentences that earn FRQ points

AP-style multiple choice

Optimization = maximize or minimize under a constraint

Objective function

Constraint

Six steps that solve every optimization problem

Classic: the open-top box

Two more: fencing and cylinders

AP-style multiple choice: optimization

Analyzing curves that aren't functions

Horizontal tangent

Vertical tangent

Increasing / decreasing

Concavity

Worked: x² + y² = 25, full analysis

Five mistakes that cost the most points in Unit 5

Two released-style MCQs

FRQ walkthrough: optimization with justification

Optimization is everywhere once you see it

📦 Packaging & logistics

💊 Drug dosage

💰 Marginal analysis in econ

🚗 Traffic enforcement

Three videos worth your time

Six problems — try before you peek

Unit 5 · complete topic coverage

Analytical Applications
of Differentiation