2023 AP Calculus AB FRQ Question 4: Analyzing a Function from its Derivative Graph

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2023.

Question 4

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Analyzing a Function from its Derivative Graph

Graph Analysis & Curve Sketching

Hard

The function

f

is defined on the closed interval

[-2,8]

and satisfies

f(2)=1

. The graph of

f'

, the derivative of

f

, consists of two line segments and a semicircle, as shown in the figure: a line segment from

(-2,2)

down to

(0,-2)

, a line segment from

(0,-2)

up to

(4,2)

, and a lower semicircle from

(4,2)

down and back up to

(8,2)

(dipping to a minimum of

0

x=6

2023 AP Calculus AB FRQ Question 4 Graph of f' — Graph of $f'$ on $[-2,8]$.

Part AMedium1 point

Does

f

have a relative minimum, a relative maximum, or neither at

x=6

? Give a reason for your answer.

Answer

Neither —

f'(x)

does not change sign at

x=6

Full Solution & Work

Check the sign of f' around x = 6

f'(x)>0

(2,6)

and

f'(x)>0

(6,8)

(the semicircle dips to touch 0 at

x=6

but never goes negative).

Conclude

f'(x)

does not change sign at

x=6

, so there is **neither** a relative maximum nor a relative minimum there.

AP Scoring — 1 Point

**P1**: Declaring

f'(x)

does not change sign at

x=6

, so neither, is sufficient — a response is not required to give specific sign intervals, but any given must be correct. Declaring

f'(x)>0

before **and** after

x=6

does not, by itself, earn this point without the "does not change sign" conclusion.

Part BMedium2 points

On what open intervals, if any, is the graph of

f

concave down? Give a reason for your answer.

Answer

(-2,0)

and

(4,6)

Full Solution & Work

Relate concavity of f to the behavior of f'

f

is concave down where

f'

is **decreasing**.

Read decreasing intervals off the graph of f'

f'

decreases on

(-2,0)

(the first line segment, negative slope) and on

(4,6)

(the left half of the semicircle, coming down from 2 to 0). So the graph of

f

is concave down on **

(-2,0)

and

(4,6)

**.

AP Scoring — 2 Points

**P1**: Earned only for the answer

(-2,0)

and

(4,6)

(endpoints may or may not be included).
**P2**: Requires P1. Must correctly discuss the behavior of

f'

(or the slopes of

f'

) as the reason.

Common mistake: A response with exactly one of the two correct intervals, with a correct reason, still earns 1 of the 2 points as a special case.

Part CHard3 points

Find the value of

\lim\limits_{x\to2} \dfrac{6f(x)-3x}{x^2-5x+6}

, or show that it does not exist. Justify your answer.

Answer

3

Full Solution & Work

Confirm the indeterminate form

Because

f

is differentiable at

x=2

f

is continuous there, so

\lim_{x\to2}f(x)=f(2)=1

\lim_{x\to2}\big(6f(x)-3x\big) = 6(1)-3(2)=0, \qquad \lim_{x\to2}(x^2-5x+6)=0

The limit is of indeterminate form

\frac{0}{0}

, so L'Hôpital's Rule can be applied.

Apply L'Hôpital's Rule

\lim_{x\to2}\frac{6f(x)-3x}{x^2-5x+6} = \lim_{x\to2}\frac{6f'(x)-3}{2x-5} = \frac{6\cdot f'(2)-3}{2(2)-5}

Read f'(2) off the graph and finish

From the graph,

f'(2)=0

(the middle line segment crosses the

x

-axis at

x=2

= \frac{6(0)-3}{4-5} = \frac{-3}{-1}=3

AP Scoring — 3 Points

**P1**: Presenting the two separate limits for the numerator and denominator (a response that presents a limit explicitly equal to

\frac{0}{0}

does not earn this point).
**P2**: Applying L'Hôpital's Rule — presenting at least one correct derivative in the limit of a ratio of derivatives.
**P3**: The correct answer, 3, with supporting work.

Part DHard3 points

Find the absolute minimum value of

f

on the closed interval

[-2,8]

. Justify your answer.

Answer

f(2)=1

Full Solution & Work

Find critical points

f'(x)=0 \implies x=-1,\ x=2,\ x=6

Evaluate f at critical points and endpoints

f

is continuous on

[-2,8]

, so the candidates for the absolute minimum are

x=-2,-1,2,6,8

\begin{array}{c|c} x & f(x) \\\hline -2 & 3 \\ -1 & 4 \\ 2 & 1 \\ 6 & 7-\pi \\ 8 & 11-2\pi\end{array}

(these are found by adding/subtracting the signed areas between consecutive

x

-values, using

f(2)=1

as the anchor value)

Conclude

The smallest value in the table is

f(2)=1

. The absolute minimum value of

f

[-2,8]

\boxed{1}

AP Scoring — 3 Points

**P1**: Stating

f'(x)=0

(or equivalent) — merely listing the zeros of

f'

is not sufficient.
**P2**: A justification:

f

is continuous on

[-2,8]

and correctly evaluated at all 5 candidates. Any error in evaluating

f

at a candidate point forfeits this point. (A response need not evaluate

f(-1)

if it argues

x=-1

is a local max via other reasoning, and need not evaluate

f(8)

if it argues

f'(x)\geq0

for

x>2

, hence

f(8)>f(2)

.)
**P3**: Earned only for indicating the minimum value is 1 — noting only that the minimum *occurs at*

x=2

is not sufficient.

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