CalcIQ · AP Calculus AB

Lesson 21: First Derivative Test & Intervals of Increase/Decrease

Unit 5 · Analytical Applications of Differentiation · Exam Weight:** 15–18% · 21/35 lessons · Mathematical Practice:** 3 — Justification

Calculator:** Mixed

Objectives:

Find the critical points of a function (where f'(x) = 0 or f'(x) is undefined and x is in the domain of f), then build a sign chart for f' to determine intervals of increase and decrease.
Apply the First Derivative Test to classify each critical point as a relative maximum, a relative minimum, or neither, and justify the classification in precise AP language.
Find absolute (global) extrema of a continuous function on a closed interval using the Candidates Test, and read increasing/decreasing and extremum information directly from the graph of f'.

(a) Opening Question

Consider the function

f(x) = x³ - 3x² - 9x + 5

Take two minutes and answer these before reading on:

Compute f'(x).
Solve f'(x) = 0. You should get two x-values. These are the only places where the tangent line is horizontal.
Pick a test value to the left of the smaller root, one between the roots, and one to the right of the larger root. Plug each into f'(x) and record only the sign (+ or −), not the value.
Wherever f' is positive, the graph is rising; wherever f' is negative, the graph is falling. Sketch a rough picture: rises, then falls, then rises. At the two turning points, which is a peak and which is a valley?

This is the entire lesson in miniature. The sign of f', not f itself, tells you where a function climbs, where it descends, and where it turns around. Everything that follows is a disciplined, AP-graded version of what you just did.

(Answers: f'(x) = 3x² - 6x - 9 = 3(x-3)(x+1); roots x = -1, 3; signs +, −, +; peak at x = -1, valley at x = 3.)

(b) Core Concepts

Critical points: where the action can happen

A critical point (also called a critical number) of f is a value x = c such that c is in the domain of f and either f'(c) = 0 or f'(c) does not exist.

Read that definition carefully — it has three pieces, and each one is an AP scoring trap:

f'(c) = 0 captures the smooth turning points, where the tangent line is horizontal.
f'(c) undefined captures corners, cusps, and vertical tangents — places where the function can still turn around even though it has no derivative there (think of the point of f(x) = |x| at x = 0).
c must be in the domain of f. If f itself is not defined at c, then c is not a critical point, no matter what f' does. For example, if f'(x) = 1/x² blows up at x = 0 but f(0) is undefined, x = 0 is not a critical point of f.

Critical points are the only candidates for relative (local) extrema in the interior of the domain. They matter because, by Fermat's Theorem, if f has a relative extremum at an interior point c where f'(c) exists, then f'(c) = 0. But the converse fails — a critical point need not be an extremum (more on this below).

Increasing and decreasing from the sign of `f'`

The link between the derivative and the shape of the graph is the foundation of Unit 5:

Increasing/Decreasing Test. Let f be continuous on an interval and differentiable on its interior.

- If f'(x) > 0 for all x in the interval, then f is increasing on that interval.

- If f'(x) < 0 for all x in the interval, then f is decreasing on that interval.

The intuition: f'(x) is the slope of the tangent line. A positive slope means the curve is going uphill as x increases; a negative slope means downhill. Between consecutive critical points, f' cannot change sign (it can only switch sign by passing through 0 or through an undefined point), so the sign of f' is constant on each such interval. That is exactly why a sign chart works.

Building a sign chart for `f'`

A sign chart for f' is the organizing tool of this lesson. Here is the procedure:

Find every critical point. Set f'(x) = 0 and solve; also find where f'(x) is undefined. Keep only the values that lie in the domain of f.
Mark these values on a number line. They divide the domain into open intervals.
Test one point inside each interval by plugging it into f' and recording the sign only.
Read the result: + means f is increasing there; − means f is decreasing there.

Mini-example. For f'(x) = 3(x-3)(x+1), the critical points are x = -1 and x = 3. Test x = -2: 3(-5)(-1) = +. Test x = 0: 3(-3)(1) = −. Test x = 4: 3(1)(5) = +.

`

x = -1 x = 3

f' + )------( − )------( +

increasing decreasing increasing

`

So f is increasing on (-∞, -1), decreasing on (-1, 3), and increasing on (3, ∞). (AP convention: report increasing/decreasing on open intervals.)

The First Derivative Test

Once the sign chart is built, classifying each critical point is automatic:

First Derivative Test. Suppose c is a critical point of a continuous function f. Look at how the sign of f' behaves as x passes through c:

- If f' changes from positive to negative at c, then f has a relative maximum at c.

- If f' changes from negative to positive at c, then f has a relative minimum at c.

- If f' does not change sign at c, then f has neither a relative max nor a relative min at c.

The justification language the AP readers want is fixed and worth memorizing word-for-word:

"f has a relative maximum at x = c because f' changes from positive to negative at x = c."

Notice that the no-change case is real. For f(x) = x³, f'(x) = 3x², which is 0 at x = 0 but is positive on both sides. The sign of f' does not change, so x = 0 is a critical point that is not an extremum — the curve flattens momentarily and keeps rising.

Absolute extrema on a closed interval: the Candidates Test

The First Derivative Test finds relative (local) extrema. To find the absolute (global) maximum and minimum of a continuous function on a closed interval [a, b], the Extreme Value Theorem guarantees both exist, and the Candidates Test finds them:

Candidates Test. If f is continuous on [a, b]:

1. Find all critical points of f in the open interval (a, b).

2. List the candidates: those critical points and the two endpoints x = a, x = b.

3. Evaluate f at every candidate.

4. The largest of these f-values is the absolute maximum; the smallest is the absolute minimum.

You do not need a sign chart for the Candidates Test — you only compare the actual f-values. The endpoints are full-fledged candidates and are the most commonly forgotten ones.

Reading the graph of `f'` (a signature AP skill)

A favorite AP setup gives you the graph of f' (not f) and asks about the behavior of f. You must translate:

What you see on the graph of `f'`	What it tells you about `f`
`f'` is above the `x`-axis (`f' > 0`)	`f` is increasing
`f'` is below the `x`-axis (`f' < 0`)	`f` is decreasing
`f'` crosses the axis from `+` to `−`	`f` has a relative maximum
`f'` crosses the axis from `−` to `+`	`f` has a relative minimum
`f'` touches the axis without crossing	critical point, but no extremum

The single biggest mistake here is treating the picture as the graph of f. On the graph of f', a peak of the curve is not a maximum of f — what matters for f's extrema is only where the f' curve crosses zero.

Reading it: f is increasing on (-2, 1) and (4, 6) because f' > 0 there, and decreasing on (1, 4) because f' < 0. f has a relative maximum at x = 1 (because f' changes from positive to negative) and a relative minimum at x = 4 (because f' changes from negative to positive).

(c) Worked Examples

Example 1 — Critical points, sign chart, and the First Derivative Test (foundation) · [NO CALC]

Problem. Let f(x) = x³ - 3x² - 9x + 5. Find all critical points, the intervals of increase and decrease, and classify each critical point. Justify with the First Derivative Test.

Strategy. Differentiate, solve f'(x) = 0, build a sign chart, apply the First Derivative Test.

Solution.

f'(x) = 3x² - 6x - 9 = 3(x² - 2x - 3) = 3(x - 3)(x + 1)

f' is a polynomial, so it is never undefined; the only critical points come from f'(x) = 0:

3(x - 3)(x + 1) = 0 ⇒ x = -1 and x = 3

Sign chart (test x = -2, x = 0, x = 4):

Interval test x sign of f' behavior of f

(-∞, -1) -2 3(-5)(-1) = + increasing

(-1, 3) 0 3(-3)(1) = − decreasing

(3, ∞) 4 3(1)(5) = + increasing

Classification (First Derivative Test).

At x = -1, f' changes from positive to negative ⇒ relative maximum.

At x = 3, f' changes from negative to positive ⇒ relative minimum.

Justification (model language). "f has a relative maximum at x = -1 because f' changes from positive to negative at x = -1, and a relative minimum at x = 3 because f' changes from negative to positive at x = 3."

(The relative max value is f(-1) = (-1) - 3(1) - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10; the relative min value is f(3) = 27 - 27 - 27 + 5 = -22. AP usually asks for the location x, so name the x-value unless the value is requested.)

Example 2 — Absolute extrema on a closed interval (procedural) · [NO CALC]

Problem. Find the absolute maximum and absolute minimum values of f(x) = x³ - 3x² - 9x + 5 on the closed interval [-2, 4].

Strategy. f is a polynomial, hence continuous on [-2, 4], so the Extreme Value Theorem applies. Use the Candidates Test.

Solution. From Example 1, the critical points are x = -1 and x = 3, both inside (-2, 4). The candidates are these two critical points plus the endpoints x = -2 and x = 4. Evaluate f at each:

f(-2) = (-8) - 3(4) - 9(-2) + 5 = -8 - 12 + 18 + 5 = 3 f(-1) = -1 - 3 + 9 + 5 = 10 f(3) = 27 - 27 - 27 + 5 = -22 f(4) = 64 - 48 - 36 + 5 = -15

candidate x f(x)

-2 (endpoint) 3

-1 (crit. pt.) 10

3 (crit. pt.) -22

4 (endpoint) -15

The largest value is 10 and the smallest is -22.

Conclusion. The absolute maximum is 10, occurring at x = -1, and the absolute minimum is -22, occurring at x = 3. (Note how the endpoint values, 3 and -15, were neither extreme here — but they had to be checked.)

Example 3 — Deducing f's behavior from the graph of f' (AP level) · [NO CALC]

Problem. The graph below is the graph of f', the derivative of a differentiable function f, on [0, 8].

(i) On what intervals is f increasing? (ii) At what x-values does f have a relative maximum or minimum? Justify. (iii) A student says, "The graph has a low point near x = 4, so f has a minimum at x = 4." Respond.

Strategy. Translate the sign of f' (above/below the axis) into the behavior of f. Ignore the height of the f' curve except for its sign.

Solution.

(i) f is increasing where f' > 0, i.e., where the graph of f' is above the axis: on (0, 2) and (6, 8). (f is decreasing on (2, 6) where f' < 0.)

(ii) Apply the First Derivative Test to the crossings of f':

At x = 2, f' changes from positive to negative ⇒ relative maximum at x = 2.

At x = 6, f' changes from negative to positive ⇒ relative minimum at x = 6.

Justification. "f has a relative maximum at x = 2 because f' changes from positive to negative there, and a relative minimum at x = 6 because f' changes from negative to positive there."

(iii) The student confused the graph of f' with the graph of f. The point near x = 4 is a low point of f', meaning f' (the slope of f) is most negative there — so f is decreasing fastest at x = 4, not bottoming out. The minimum of f is at x = 6, where f' crosses from negative to positive.

Example 4 — A critical point where f' is undefined (AP level) · [NO CALC]

Problem. Let f(x) = x^(2/3). Find all critical points and classify any relative extrema, justifying your answer.

Strategy. Differentiate, then check both conditions for critical points: f'(x) = 0 and f'(x) undefined while x is in the domain.

Solution. The domain of f(x) = x^(2/3) is all real numbers (the cube root is defined for negatives). Differentiate:

f'(x) = (2/3) x^(-1/3) = 2 / (3·∛x)

f'(x) = 0 has no solution — the numerator 2 is never zero.

f'(x) is undefined at x = 0 (division by ∛0 = 0). Since x = 0 is in the domain of f (indeed f(0) = 0), x = 0 is a critical point.

Sign chart (test x = -1 and x = 1):

f'(-1) = 2 / (3·(-1)) = -2/3 < 0 ⇒ f decreasing on (-∞, 0) f'(1) = 2 / (3·(1)) = +2/3 > 0 ⇒ f increasing on (0, ∞)

At x = 0, f' changes from negative to positive.

Conclusion. f has a relative (and absolute) minimum at x = 0, with value f(0) = 0.

Justification. "x = 0 is a critical point because f'(0) does not exist while 0 is in the domain of f. f has a relative minimum at x = 0 because f' changes from negative to positive at x = 0." (Graphically, f(x) = x^(2/3) has a sharp cusp at the origin — a turning point with no tangent line.)

(d) Common Mistakes

Building the sign chart from f instead of f'.

What students do: Plug test values into the original function f and use the sign of f to decide increasing/decreasing.

Why it's wrong: The sign of f tells you whether the curve is above or below the x-axis, not whether it is rising or falling. Increase/decrease is governed entirely by f'.

Fix: The sign chart is always a chart of f'. Test values go into the derivative.

Forgetting critical points where f' is undefined (or ignoring the domain).

What students do: Only solve f'(x) = 0 and stop, missing corners, cusps, and vertical tangents — or they include a value where f' is undefined even though f itself is undefined there.

Why it's wrong: Critical points include places where f' does not exist, but only if x is in the domain of f. Both halves of the definition matter.

Fix: For every problem, ask two questions: "Where is f'(x) = 0?" and "Where is f'(x) undefined while f is still defined?"

Calling every critical point an extremum.

What students do: Assume that wherever f'(x) = 0, there is automatically a max or min.

Why it's wrong: If f' does not change sign at c (as with f(x) = x³ at x = 0), there is no extremum. A horizontal tangent alone is not enough.

Fix: Always check the sign change of f' across the critical point. No sign change ⇒ neither a max nor a min.

Confusing the graph of f' with the graph of f.

What students do: On a "graph of f'" problem, read peaks and valleys of the picture as maxima and minima of f.

Why it's wrong: On the graph of f', the extrema of f occur where the curve crosses the x-axis, not at the high and low points of the f' curve (those locate where f changes concavity — a Lesson 22 idea).

Fix: On a graph of f', watch the sign of the curve (above vs. below the axis) and its zero crossings. Write "this is the graph of f'" at the top of your scratch work to stay oriented.

Forgetting the endpoints in the Candidates Test.

What students do: Find critical points on [a, b], evaluate f there, and pick the largest/smallest — without checking f(a) and f(b).

Why it's wrong: On a closed interval, an absolute extremum can occur at an endpoint. Skipping endpoints can miss the true max or min entirely.

Fix: The candidate list is always critical points in (a,b) plus both endpoints. Evaluate f at all of them.

(e) Practice Problems

For multiple choice, choose the single best answer. Calculator labels match AP rules.

Question 1NO CALC

The critical points of f(x) = x³ - 12x are:

(A) x = -2 and x = 2

(B) x = -4 and x = 4

(C) x = 0 only

(D) x = 12

Question 2NO CALC

On which interval is f(x) = x³ - 12x decreasing?

(A) (2, ∞)

(B) (-2, 2)

(C) (-∞, -2)

(D) (-∞, ∞)

Question 3NO CALC

Let g(x) = x⁴ - 4x³. The function g is increasing on:

(A) (0, 3)

(B) (-∞, 3)

(C) (-∞, 0)

(D) (3, ∞)

Question 4NO CALC

If f'(x) = (x - 1)²(x + 4), then f has:

(A) relative extrema at both x = 1 and x = -4

(B) a relative max at x = 1 and a relative min at x = -4

(C) a relative max at x = -4 only

(D) a relative min at x = -4 only

Question 5NO CALC

A function has f'(x) = x²(x - 2). At x = 0, f has:

(A) neither a max nor a min

(B) a relative maximum

(C) cannot be determined

(D) a relative minimum

Question 6CALC

For f(x) = x³ - 4x² + x + 6 on [0, 4], the absolute minimum value of f occurs at x ≈

(A) 0

(B) 0.131

(C) 2.535

(D) 4

Question 7NO CALC

The function f(x) = x²ᐟ³(x - 5) has a critical point at x = 0 because:

(A) f(0) = 0

(B) f'(0) does not exist and 0 is in the domain of f

(C) f'(0) = 0

(D) x = 0 is an endpoint

Question 8NO CALC

(Graph of f'.) The graph of f', the derivative of f, is positive on (-∞, -1), negative on (-1, 3), and positive on (3, ∞). Then f has:

(A) a relative min at x = -1, relative max at x = 3

(B) no relative extrema

(C) relative maxima at both x = -1 and x = 3

(D) a relative max at x = -1, relative min at x = 3

Question 9CALC

Let f(x) = x + 2sin(x) on [0, 2π]. The number of critical points of f in the open interval (0, 2π) is:

(A) 0

(B) 1

(C) 2

(D) 3

Question 10NO CALC

If f is continuous on [1, 5] with critical points at x = 2 and x = 4, and f(1) = 3, f(2) = 7, f(4) = -1, f(5) = 5, then the absolute maximum value of f on [1, 5] is:

(A) -1

(B) 7

(C) 5

(D) 3

Question 11NO CALC

The graph of f' lies entirely below the x-axis on (a, b) except that it touches the axis (without crossing) at a single point x = c in (a, b). On (a, b), the function f is:

(A) constant

(B) increasing then decreasing

(C) decreasing on all of (a, b)

(D) increasing on all of (a, b)

(Short answer)Let f(x) = 3x⁴ - 8x³ + 6. Find all critical points, the intervals of increase and decrease, and classify each critical point using the First Derivative Test.

(Short answer)Find the absolute maximum and absolute minimum values of f(x) = 2x³ - 3x² - 12x + 1 on the closed interval [-2, 3]. Show the candidates.

(Justification)(Graph of f'.) The graph of f' on [0, 6] is below the x-axis on (0, 2), crosses upward at x = 2, is above the axis on (2, 5), and crosses downward at x = 5. State where f is increasing, locate every relative extremum of f, and justify each classification with proper First Derivative Test language.

(Justification)A student claims, "Since f'(3) = 0, the function f(x) = (x - 3)³ + 4 must have a relative extremum at x = 3." Determine whether the claim is correct, and justify your answer using the sign of f'.

(f) AP Exam Focus

FRQ — Analysis from the Graph of f' (Section II style). Calculator NOT permitted. Total: 9 points.

Let f be a function that is continuous on the closed interval [-3, 5] and differentiable on the open interval (-3, 5). The graph of f', the derivative of f, is shown below. It consists of straight line segments and passes through the points indicated.

Additionally, you are told that f(-3) = 6.

(a) Find all open intervals on which f is increasing and all open intervals on which f is decreasing. Justify your answer. (3 points)

(b) At what value(s) of x in the open interval (-3, 5) does f have a relative maximum? At what value(s) does f have a relative minimum? Justify your answers. (3 points)

(c) Explain why f does not have a relative extremum at x = 1, even though x = 1 is a notable feature of the graph of f'. (1 point)

(d) The absolute maximum of f on [-3, 5] occurs at one of the candidates x = -3, x = -1, or x = 5. Identify which x-value the absolute maximum candidate analysis must compare, and explain how you would determine the absolute maximum. (2 points)

Model Solution

(a) f is increasing where f' > 0 and decreasing where f' < 0. Reading the sign of the graph of f':

f' > 0 on (-3, -1) and on (3, 5) ⇒ f is increasing there f' < 0 on (-1, 3) ⇒ f is decreasing there

f is increasing on (-3, -1) and (3, 5) because f' > 0 on those intervals, and decreasing on (-1, 3) because f' < 0 there.

(b) Apply the First Derivative Test at the zeros of f' where the sign changes:

At x = -1, f' changes from positive to negative ⇒ relative maximum at x = -1.

At x = 3, f' changes from negative to positive ⇒ relative minimum at x = 3.

Justification. "f has a relative maximum at x = -1 because f' changes from positive to negative at x = -1. f has a relative minimum at x = 3 because f' changes from negative to positive at x = 3."

(c) At x = 1, the graph of f' reaches its lowest point, but f'(1) = -2 ≠ 0, so x = 1 is not a critical point of f. The sign of f' is negative on both sides of x = 1, so f' does not change sign there. Therefore f has no relative extremum at x = 1; in fact f is simply decreasing through x = 1 (and decreasing fastest there, since f' is most negative).

(d) Because f is increasing on (-3, -1), then decreasing on (-1, 3), then increasing on (3, 5), the absolute maximum on [-3, 5] must be at the left endpoint x = -3, the relative maximum x = -1, or the right endpoint x = 5 (the relative minimum at x = 3 cannot be the maximum). To determine the absolute maximum, evaluate f at x = -3, x = -1, and x = 5, and compare the values; the largest is the absolute maximum. (Concretely, f(-1) = f(-3) + ∫₋₃⁻¹ f'(x)dx > f(-3) since f' > 0 on (-3,-1), so x = -3 is eliminated and the comparison is between x = -1 and x = 5.)

Scoring Commentary — where students lose points

Part (a) [3 pts]: 1 pt for the correct increasing intervals (-3, -1) ∪ (3, 5); 1 pt for the correct decreasing interval (-1, 3); 1 pt for the justification that references f' > 0/f' < 0. Students who give intervals but never say why (the sign of f') lose the third point. Using the graph as if it were f is the classic zero-credit error.

Part (b) [3 pts]: 1 pt for the relative max at x = -1; 1 pt for the relative min at x = 3; 1 pt for First Derivative Test justification ("f' changes from positive to negative / negative to positive"). A bare answer with no sign-change language earns the location points but not the justification point.

Part (c) [1 pt]: Full credit requires noting that f'(1) ≠ 0 (or that f' does not change sign at x = 1), so x = 1 is not even a critical point. Saying only "it's a minimum of the graph" without connecting to the sign of f' earns nothing.

Part (d) [2 pts]: 1 pt for identifying that the comparison is among the endpoints and the relative maximum (x = -3, -1, 5); 1 pt for the method ("evaluate f and take the largest"). A student who tries to find a max at the relative minimum x = 3, or who forgets the endpoints, loses credit.

Justification language that earns full credit: "f has a relative maximum at x = -1 because f' changes from positive to negative at x = -1." Vague phrasing such as "the graph turns around" earns no justification credit.

🔑 Answer Key

1. (A). f'(x) = 3x² - 12 = 3(x² - 4) = 3(x-2)(x+2) = 0 ⇒ x = ±2. Distractors: (C) sets only x = 0; (B) mis-solves x² = 16; (D) reads off a coefficient.

2. (B). f'(x) = 3(x-2)(x+2); f' < 0 between the roots, on (-2, 2). Distractors: (C)/(A) are the increasing intervals; (D) ignores that f' changes sign.

3. (D). g'(x) = 4x³ - 12x² = 4x²(x - 3); critical points x = 0, 3. Sign of g': negative on (-∞,0) and (0,3) (since 4x² ≥ 0 and (x-3) < 0), positive on (3, ∞). So g increases only on (3, ∞). Distractors: (A)/(C)/(B) come from forgetting that 4x² does not change the sign and that g' stays negative through x = 0.

4. (D). f'(x) = (x-1)²(x+4). At x = -4, f' changes from − to + (since (x-1)² > 0, the sign follows (x+4)) ⇒ relative min. At x = 1, (x-1)² keeps f' the same sign on both sides (no sign change) ⇒ neither. Distractors: (B)/(C)/(A) wrongly treat the repeated factor x = 1 as an extremum.

5. (A). f'(x) = x²(x-2). At x = 0, the factor x² does not change sign, so f' is negative on both sides of 0 (since x - 2 < 0 near 0) — no sign change ⇒ neither. Distractors: (B)/(D) assume f'(0)=0 forces an extremum; (C) is determinable.

6. (C). f'(x) = 3x² - 8x + 1 = 0 ⇒ x = (8 ± √(64-12))/6 = (8 ± √52)/6 ≈ 0.131, 2.535. Candidates on [0,4]: f(0)=6, f(0.131)≈6.065, f(2.535)≈ -0.879, f(4)=10. The minimum is at x ≈ 2.535. Distractors: (A)/(D) are endpoints (not minimal); (B) is the relative max location.

7. (B). f(x) = x²ᐟ³(x-5) = x⁵ᐟ³ - 5x²ᐟ³, so f'(x) = (5/3)x²ᐟ³ - (10/3)x⁻¹ᐟ³, which is undefined at x = 0; since 0 is in the domain of f, x = 0 is a critical point. Distractors: (C) is false (f'(0) does not exist, it is not 0); (A) f(0)=0 alone is irrelevant; (D) x=0 is interior, not an endpoint.

8. (D). f' > 0 then < 0 at x = -1 ⇒ relative max; f' < 0 then > 0 at x = 3 ⇒ relative min. Distractors: (A) swaps max/min; (C)/(B) misread the sign changes.

9. (C). f'(x) = 1 + 2cos(x) = 0 ⇒ cos(x) = -1/2 ⇒ x = 2π/3 ≈ 2.094 and x = 4π/3 ≈ 4.189, both in (0, 2π). So 2 critical points. Distractors: (B)/(D) miscount the solutions of cos x = -1/2 on (0, 2π).

10. (B). Compare f at all candidates: f(1)=3, f(2)=7, f(4)=-1, f(5)=5. Largest is 7 at x = 2. Distractors: (D)/(C) are non-maximal endpoint values; (A) is the minimum.

11. (C). f' ≤ 0 throughout, and the single touch point (where f' = 0 without a sign change) does not interrupt the decrease. So f is decreasing on all of (a, b). Distractors: (D)/(B) misread a non-crossing touch as a turning point; (A) constant would require f' ≡ 0.

12. f'(x) = 12x³ - 24x² = 12x²(x - 2); critical points x = 0 and x = 2.

Sign of f': on (-∞, 0), 12x² > 0 and (x-2) < 0 ⇒ f' < 0 (decreasing); on (0, 2), still (x-2) < 0 ⇒ f' < 0 (decreasing); on (2, ∞), f' > 0 (increasing).

- At x = 0: f' does not change sign (negative on both sides) ⇒ neither a max nor a min.

- At x = 2: f' changes from negative to positive ⇒ relative minimum.

Intervals: f is decreasing on (-∞, 2) and increasing on (2, ∞). Justification: "f has a relative minimum at x = 2 because f' changes from negative to positive at x = 2; at x = 0, f' does not change sign, so f has no extremum there."

13. f'(x) = 6x² - 6x - 12 = 6(x² - x - 2) = 6(x - 2)(x + 1); critical points x = -1, 2, both in (-2, 3). Candidates and values:

f(-2) = 2(-8) - 3(4) - 12(-2) + 1 = -16 - 12 + 24 + 1 = -3;

f(-1) = 2(-1) - 3(1) - 12(-1) + 1 = -2 - 3 + 12 + 1 = 8;

f(2) = 2(8) - 3(4) - 12(2) + 1 = 16 - 12 - 24 + 1 = -19;

f(3) = 2(27) - 3(9) - 12(3) + 1 = 54 - 27 - 36 + 1 = -8.

Absolute maximum = 8 at x = -1; absolute minimum = -19 at x = 2.

14. f is increasing where f' > 0: on (2, 5). f is decreasing where f' < 0: on (0, 2) and (5, 6).

- At x = 2, f' changes from negative to positive ⇒ relative minimum at x = 2.

- At x = 5, f' changes from positive to negative ⇒ relative maximum at x = 5.

Justification: "f has a relative minimum at x = 2 because f' changes from negative to positive at x = 2, and a relative maximum at x = 5 because f' changes from positive to negative at x = 5."

15. The claim is incorrect. f'(x) = 3(x - 3)², which is 0 at x = 3 but is positive on both sides of x = 3 (a square is nonnegative). Since f' does not change sign at x = 3, the First Derivative Test gives neither a maximum nor a minimum there — f is increasing through x = 3 with a momentary horizontal tangent. A horizontal tangent (f' = 0) is necessary but not sufficient for a relative extremum; the sign of f' must change.

CalcIQ · Lesson 21 of 35 · Unit 5 — Analytical Applications of Differentiation. Next: Lesson 22 — Second Derivative Test & Concavity.

This lesson is independent study material and is not endorsed by or affiliated with the College Board. "AP" and "Advanced Placement" are trademarks of the College Board.

Accuracy note: All derivatives, critical points, sign charts, and First Derivative Test classifications in this lesson were independently recomputed and verified. Each relative extremum is justified by an explicit change in the sign of f', and all absolute extrema were found by evaluating f at every candidate (critical points and endpoints) on the closed interval.

← Lesson 20

Lesson 22 →

CalcIQ · Aligned to the AP Calculus AB framework (8 units). Not affiliated with the College Board. AP® is a registered trademark of the College Board. Content pending mathematical-accuracy review.

Lesson 21: First Derivative Test & Intervals of Increase/Decrease

(a) Opening Question

(b) Core Concepts

Critical points: where the action can happen

Increasing and decreasing from the sign of `f'`

Building a sign chart for `f'`

The First Derivative Test

Absolute extrema on a closed interval: the Candidates Test

Reading the graph of `f'` (a signature AP skill)

(c) Worked Examples

Example 1 — Critical points, sign chart, and the First Derivative Test (foundation) · [NO CALC]

Example 2 — Absolute extrema on a closed interval (procedural) · [NO CALC]

Example 3 — Deducing `f`'s behavior from the graph of `f'` (AP level) · [NO CALC]

Example 4 — A critical point where `f'` is undefined (AP level) · [NO CALC]

(d) Common Mistakes

(e) Practice Problems

(f) AP Exam Focus

Model Solution

Scoring Commentary — where students lose points

Interval	test `x`	sign of `f'`	behavior of `f`
`(-∞, -1)`	`-2`	`3(-5)(-1) = +`	increasing
`(-1, 3)`	`0`	`3(-3)(1) = −`	decreasing
`(3, ∞)`	`4`	`3(1)(5) = +`	increasing

candidate `x`	`f(x)`
`-2` (endpoint)	`3`
`-1` (crit. pt.)	`10`
`3` (crit. pt.)	`-22`
`4` (endpoint)	`-15`

Lesson 21: First Derivative Test & Intervals of Increase/Decrease

(a) Opening Question

(b) Core Concepts

Critical points: where the action can happen

Increasing and decreasing from the sign of f'

Building a sign chart for f'

The First Derivative Test

Absolute extrema on a closed interval: the Candidates Test

Reading the graph of f' (a signature AP skill)

(c) Worked Examples

Example 1 — Critical points, sign chart, and the First Derivative Test (foundation) · [NO CALC]

Example 2 — Absolute extrema on a closed interval (procedural) · [NO CALC]

Example 3 — Deducing f's behavior from the graph of f' (AP level) · [NO CALC]

Example 4 — A critical point where f' is undefined (AP level) · [NO CALC]

(d) Common Mistakes

(e) Practice Problems

(f) AP Exam Focus

Model Solution

Scoring Commentary — where students lose points

Increasing and decreasing from the sign of `f'`

Building a sign chart for `f'`

Reading the graph of `f'` (a signature AP skill)

Example 3 — Deducing `f`'s behavior from the graph of `f'` (AP level) · [NO CALC]

Example 4 — A critical point where `f'` is undefined (AP level) · [NO CALC]