2022 AP Calculus AB FRQ Question 3: Analyzing a Function from its Derivative Graph

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2022.

Question 3

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Analyzing a Function from its Derivative Graph

Graph Analysis & Curve Sketching

Hard

Let

f

be a differentiable function with

f(4)=3

. On the interval

0\leq x\leq7

, the graph of

f'

, the derivative of

f

, consists of a semicircle (radius 2, centered at

(2,0)

, dipping below the axis) and two line segments: from

(4,0)

(6,2)

, and from

(6,2)

(7,1)

2022 AP Calculus AB FRQ Question 3 Graph of f' — Graph of $f'$ on $[0,7]$.

Part AMedium2 points

Find

f(0)

and

f(5)

Answer

f(0) = 3+2\pi

;

f(5)=3.5

Full Solution & Work

Find f(0) using the semicircle's signed area

f(0) = f(4) - \int_0^4 f'(x)\,dx = 3 - \left(-\frac12\pi(2)^2\right) = 3+2\pi

Find f(5) using the triangle's area

f(5) = f(4)+\int_4^5 f'(x)\,dx = 3+\frac12(1)(1) = 3.5

AP Scoring — 2 Points

**P1**:

f(0)=3+2\pi

.
**P2**:

f(5)=3.5

Part BHard2 points

Find the

x

-coordinates of all points of inflection of the graph of

f

for

0<x<7

. Justify your answer.

Answer

x=2

and

x=6

Full Solution & Work

Relate inflection points to local extrema of f'

f

has a point of inflection where

f''=(f')'

changes sign — i.e., where

f'

has a local min or max.

Locate local extrema of f'

f'

has a local minimum at

x=2

(bottom of the semicircle) and a local maximum at

x=6

(peak of the "tent" formed by the two line segments).

Conclude

f

has points of inflection at

x=2

and

x=6

(at

x=4

f'

keeps increasing through the corner, so there's no sign change and no inflection point there).

AP Scoring — 2 Points

**P1**: Correct values

x=2

and

x=6

(no additional/incorrect values).
**P2**: Justification tied to

f'

having a local extremum (or

f''

changing sign) at each location.

Part CHard3 points

Let

g

be the function defined by

g(x)=f(x)-x

. On what intervals, if any, is

g

decreasing for

0\leq x\leq7

? Show the analysis that leads to your answer.

Answer

(0,5)

Full Solution & Work

Set up g'(x) = f'(x) - 1

g

is decreasing where

g'(x)=f'(x)-1<0

, i.e. where

f'(x)<1

Compare f'(x) to 1 on each piece

[0,4]

(the semicircle):

f'(x)\leq0<1

throughout.
On

[4,6]

(line, slope 1, from

(4,0)

(6,2)

f'(x)=x-4<1

for

x<5

.
On

[6,7]

(line from

(6,2)

(7,1)

f'(x)=8-x\geq1

throughout, so

f'\geq1

here.

Conclude

f'(x)<1

(so

g

is decreasing) precisely on

(0,5)

AP Scoring — 3 Points

**P1**: Setting up

g'(x)=f'(x)-1

.
**P2**: Correctly comparing

f'(x)

to 1 on each piece of the domain.
**P3**: Correct interval,

(0,5)

, with supporting analysis.

Part DHard2 points

For the function

g

defined in part (c), find the absolute minimum value on the interval

0\leq x\leq7

. Justify your answer.

Answer

-1.5

, at

x=5

Full Solution & Work

Identify the critical point

From part (c),

g

decreases on

(0,5)

and increases on

(5,7)

(since

f'(x)\geq1

throughout

(5,7)

), so

x=5

is the location of the global minimum on

[0,7]

Evaluate g(5)

g(5) = f(5)-5 = 3.5-5=-1.5

Confirm against the endpoints

g(0)=f(0)-0=3+2\pi\approx9.283

and

g(7)=f(7)-7

, where

f(7)=f(6)+\int_6^7f'(x)\,dx=5+1.5=6.5

, so

g(7)=-0.5

. Both exceed

-1.5

, confirming the minimum is at

x=5

AP Scoring — 2 Points

**P1**: Correctly identifies

x=5

as the global minimum location, with justification (sign change of

g'

from negative to positive).
**P2**: Correct value,

g(5)=-1.5

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