Free Response · Interpreting the derivative to understand the original function

[2026] Graph Analysis AP Calculus AB FRQ: f'(x) Graphs & FTC

You are typically given the graph of $f'(x)$ (the derivative) and an initial condition $f(a) = k$ . You must use the Area Under the Curve to find values of $f(x)$ , identify local extrema, and determine concavity/points of inflection.

The Fundamental Theorem of Calculus (Accumulation) $f(x) = f(a) + \int_a^x f'(t)\,dt$
When to useTo find a specific value $f(x)$ , start at the known point $f(a)$ and add the 'net area' from $a$ to $x$ .
Area as Net Change $\int_a^b f'(x)\,dx = f(b) - f(a)$
When to useThe definite integral is the area between the $f'(x)$ graph and the x-axis. Areas above the axis are positive; areas below are negative.
First Derivative Test (Extrema) $f'(x) = 0 \text{ or undefined}$
When to useCritical points occur where the graph of $f'$ crosses or touches the x-axis. $f'$ changes $+$ to $-$ (Relative Max); $f'$ changes $-$ to $+$ (Relative Min).
Concavity and $f''(x)$ $f''(x) = \frac{d}{dx}[f'(x)]$
When to useThe slope of the $f'(x)$ graph is $f''(x)$ . If $f'(x)$ is increasing (positive slope), $f(x)$ is concave up.

Step-by-Step SOP

1
Identify the 'Home Base'
Locate the given point $(a, f(a))$ . This is your starting constant for all FTC calculations. If you go backwards (e.g., $\int_5^2$ ), remember to flip the sign of the area.
2
Map out $f(x)$ 's Behavior
Scan the $f'$ graph: Where is it above the x-axis ( $f$ increases)? Where is it below ( $f$ decreases)? Where are the 'peaks and valleys' of $f'$ (Inflection points of $f$ )?
3
Check the Endpoints for Absolute Extrema
If asked for the 'Absolute Maximum/Minimum' on $[a, b]$ , you MUST test the critical points AND the endpoints $a$ and $b$ . This is the Candidates Test.
4
Justify with Calculus Terms
Never say 'the graph goes up' or 'it changes signs.' Always say ' $f'(x)$ changes from positive to negative' or ' $f'(x)$ is increasing.'

Practice Questions

1 2 3

Question 1

2025 AP Calculus AB Exam — FRQ 4

Medium

The continuous function

f

is defined on the closed interval

-6 \le x \le 12

. The graph of

f

, consisting of two semicircles and one line segment, is shown in the figure. Let

g

be the function defined by

g(x) = \int_{6}^{x} f(t) \, dt

Graph of f consisting of two semicircles and one line segment — The function f is defined on [-6, 12]. g(x) is the integral of f from 6 to x.

APart A

Easy

Find

g'(8)

. Give a reason for your answer.

Need a Hint?

By the Fundamental Theorem of Calculus, if

g(x) = \int_{6}^{x} f(t) \, dt

, then

g'(x) = f(x)

. Find the value of

f(8)

from the graph.

Show Solution

Apply FTC Part 1

g'(x) = \frac{d}{dx} \int_{6}^{x} f(t) \, dt = f(x)

Identify f(8) from the graph

From the graph, the line segment passes through

(6, 0)

and

(12, 3)

. At

x = 8

f(8) = 1

AP Scoring Note — P1 + P2

P1 (Reason): Earned for explicitly stating

g'(x) = f(x)

g'(8) = f(8)

.

P2 (Answer):

g'(8) = 1

. Note: Simply writing

g'(8) = 1

without mentioning

f(8)

might lose the reasoning point depending on the year's stringency.

BPart B

Medium

Find all values of

x

in the open interval

-6 < x < 12

at which the graph of

g

has a point of inflection. Give a reason for your answer.

Need a Hint?

A point of inflection for

g

occurs when

g''(x)

changes sign. Since

g'(x) = f(x)

g''(x) = f'(x)

. Look for where

f(x)

changes from increasing to decreasing or vice versa.

Show Solution

Identify POI criteria

Inflection points of

g

occur where

g'(x) = f(x)

changes from increasing to decreasing (relative extrema of

f

Locate points on graph

f

changes direction at

x = -3

(dec to inc),

x = 3

(inc to dec), and

x = 6

(dec to inc).

AP Scoring Note — P3 + P4

P3 (Answer): Must list all three:

x = -3, 3, 6

.

P4 (Reason): You must tie the reason to the graph of

f

. Correct: '

f

changes from increasing to decreasing or vice versa'. Incorrect: 'The slope of the graph changes' (too vague).

CPart C

Medium

Find

g(12)

and

g(0)

. Label your answers.

Need a Hint?

Use geometry to find the area under the curve. Remember

g(0) = \int_{6}^{0} f(t) \, dt = -\int_{0}^{6} f(t) \, dt

Show Solution

Calculate g(12)

g(12) = \int_{6}^{12} f(t) \, dt = \frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(6)(3) = 9

Calculate g(0)

g(0) = \int_{6}^{0} f(t) \, dt = -\int_{0}^{6} f(t) \, dt

. The area from 0 to 6 is a semicircle with

r=3

\frac{1}{2}\pi(3)^2 = \frac{9\pi}{2}

. Therefore,

g(0) = -\frac{9\pi}{2}

AP Scoring Note — P5 + P6

P5 & P6 (Answers): Unlabeled values (just numbers without 'g(12)=') will earn 0 points.

You do NOT need to simplify

\frac{1}{2}(6)(3)

to 9 to get the point.

DPart D

Hard

Find the value of

x

at which

g

attains an absolute minimum on

[-6, 12]

. Justify your answer.

Need a Hint?

Use the Candidates Test. Check the endpoints (

x = -6, 12

) and critical points where

g'(x) = f(x) = 0

Show Solution

Identify Candidates

Critical points (

f(x)=0

x = 0

and

x = 6

. Endpoints:

x = -6

and

x = 12

Evaluate g(x) at each candidate

g(-6) = \int_{6}^{-6} f(t) \, dt = -(\text{Area}_1 + \text{Area}_2) = -(\frac{9\pi}{2} - \frac{9\pi}{2}) = 0

g(0) = -\frac{9\pi}{2} \approx -14.137

g(6) = 0

g(12) = 9

Compare and Conclude

The smallest value is

-\frac{9\pi}{2}

x = 0

AP Scoring Note — P7 + P8 + P9

P7 (Critical Points): Earned for considering where

f(x) = 0

.

P8 (Justification): You must show a complete Candidates Test (table or list) or a global sign-change argument.

P9 (Answer):

x = 0

. Note: If you only use a local argument (First Derivative Test), you lose the justification point.

Question 2

2024 AP Calculus AB Exam — FRQ 4

Medium

The graph of the differentiable function

f

, shown for

-6 \leq x \leq 7

, has a horizontal tangent at

x = -2

and is linear for

0 \le x \le 7

. Let

R

be the region in the second quadrant bounded by the graph of

f

, the vertical line

x = -6

, and the

x

- and

y

-axes. Region

R

has area 12.

Graph of f with shaded region R — The graph of f is differentiable for [-6, 7] and linear for [0, 7]. Region R has area 12.

APart A

Easy

The function

g

is defined by

g(x) = \int_{0}^{x} f(t) \, dt

. Find the values of

g(-6)

g(4)

, and

g(6)

Need a Hint?

Remember that

\int_a^b f(t)dt = -\int_b^a f(t)dt

. Use geometric areas for the linear parts and the given area for region R.

Show Solution

Calculate g(-6)

g(-6) = \int_{0}^{-6} f(t) \, dt = -\int_{-6}^{0} f(t) \, dt = -(\text{Area of } R) = -12

Calculate g(4)

g(4) = \int_{0}^{4} f(t) \, dt

. This is a triangle with base 4 and height 2. Area =

\frac{1}{2}(4)(2) = 4

Calculate g(6)

g(6) = g(4) + \int_{4}^{6} f(t) \, dt = 4 + \frac{1}{2}(2)(-1) = 4 - 1 = 3

AP Scoring Note — P1, P2, P3

P1, P2, P3: Each correct value earns 1 point. Supporting work is not required, but if shown, it must be correct. Unlabeled values are read in order.

BPart B

Easy

For the function

g

defined in part (a), find all values of

x

in the interval

0 \le x \le 6

at which the graph of

g

has a critical point. Give a reason for your answer.

Need a Hint?

A critical point of

g

occurs where

g'(x) = 0

or is undefined. Use FTC to relate

g'

f

Show Solution

Identify Relationship

By FTC,

g'(x) = f(x)

Solve for Critical Points

g'(x) = f(x) = 0 \implies x = 4

(within the interval

[0, 6]

AP Scoring Note — P4, P5

P4 (FTC): Earned for explicitly writing

g'(x) = f(x)

.

P5 (Answer+Reason):

x = 4

with the reason that

g'(4) = 0

. Reporting extra points in the interval loses this point.

CPart C

Medium

The function

h

is defined by

h(x) = \int_{-6}^{x} f'(t) \, dt

. Find the values of

h(6)

h'(6)

, and

h''(6)

. Show the work that leads to your answers.

Need a Hint?

For

h(6)

, use FTC:

\int_a^b f'(t)dt = f(b) - f(a)

. For

h'

and

h''

, differentiate

h(x)

and use the properties of the graph of f.

Show Solution

Calculate h(6)

h(6) = \int_{-6}^{6} f'(t) \, dt = f(6) - f(-6) = -1 - 0.5 = -1.5

Calculate h'(6)

h'(x) = \frac{d}{dx} \int_{-6}^{x} f'(t) \, dt = f'(x)

. Since

f

is linear on

[0, 7]

f'(6)

is the slope of the line:

\frac{-1 - 2}{6 - 0} = -\frac{3}{6} = -\frac{1}{2}

. So

h'(6) = -0.5

Calculate h''(6)

h''(x) = f''(x)

. Since

f

is linear on

[0, 7]

, its second derivative is

0

. So

h''(6) = 0

AP Scoring Note — P6, P7, P8, P9

P6 (FTC): Use of

f(6)-f(-6)

.

P7 (h(6) Value): Correct calculation of

-1.5

.

P8 (h'(6)): Earned for stating

h'(6) = f'(6)

and the value

-1/2

.

P9 (h''(6)): Earned for

0

Question 3

2023 AP Calculus AB Exam — FRQ 4

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.

Graph of f' consisting of two line segments and a semicircle — The derivative function f' is defined on [-2, 8]. f(2) = 1 is given.

APart A

Easy

Does

f

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

x = 6

? Give a reason for your answer.

Need a Hint?

Relative extrema occur where

f'(x)

changes sign. Check the graph of

f'

x=6

Show Solution

Analyze the sign of f'(x)

From the graph,

f'(x) > 0

on the interval

(2, 6)

and

f'(x) > 0

on the interval

(6, 8)

Determine Extrema

Since

f'(x)

does not change sign at

x = 6

, there is neither a relative maximum nor a relative minimum at this location.

AP Scoring Note — P1

P1 (Answer w/ Reason): You must explicitly state that

f'

'does not change sign'. Simply saying

f'(6)=0

is not enough.

BPart B

Medium

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

f

concave down? Give a reason for your answer.

Need a Hint?

A function is concave down when

f''(x) < 0

, which means

f'(x)

is decreasing.

Show Solution

Identify where f' is decreasing

Looking at the graph,

f'(x)

is decreasing on the intervals

(-2, 0)

and

(4, 6)

State the reason

The graph of

f

is concave down on

(-2, 0)

and

(4, 6)

because

f'

is decreasing on these intervals.

AP Scoring Note — P2 + P3

P2 (Intervals): Earned for the correct intervals (inclusion of endpoints is okay).

P3 (Reason): You must link concavity to the behavior of

f'

(decreasing). Mentioning

f'' < 0

alone may not be enough if not tied to the given graph of

f'

CPart C

Medium

Find the value of

\lim_{x \to 2} \frac{6f(x) - 3x}{x^2 - 5x + 6}

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

Need a Hint?

First check the limit of the numerator and denominator separately. If you get 0/0, apply L'Hospital's Rule.

Show Solution

Test for Indeterminate Form

Since

f

is differentiable, it is continuous.

\lim_{x \to 2} (6f(x) - 3x) = 6(1) - 3(2) = 0

\lim_{x \to 2} (x^2 - 5x + 6) = 2^2 - 5(2) + 6 = 0

Apply L'Hospital's Rule

Using L'Hospital's Rule:

\lim_{x \to 2} \frac{6f'(x) - 3}{2x - 5}

Evaluate the Limit

From the graph,

f'(2) = 0

.
Limit =

\frac{6(0) - 3}{2(2) - 5} = \frac{-3}{-1} = 3

AP Scoring Note — P4 + P5 + P6

P4 (Limits): You MUST show the limits of the numerator and denominator separately equaling 0. Writing '0/0' as an equality will lose this point.

P5 (L'Hospital's): Earned for correctly differentiating the top and bottom.

P6 (Answer): 3.

DPart D

Hard

Find the absolute minimum value of

f

on the closed interval

[-2, 8]

. Justify your answer.

Need a Hint?

Use the Candidates Test: Check endpoints

x = -2, 8

and critical points where

f'(x) = 0

Show Solution

Identify Critical Points

f'(x) = 0

x = -1, 2, 6

Evaluate candidates using FTC

f(2) = 1

(Given).

f(-2) = f(2) + \int_{2}^{-2} f'(t) \, dt = 1 - (\text{net area from -2 to 2}) = 1 - (-2) = 3

f(6) = f(2) + \int_{2}^{6} f'(t) \, dt = 1 + (\text{area of triangle}) = 1 + \frac{1}{2}(4)(2) = 5

f(8) = f(6) + \int_{6}^{8} f'(t) \, dt = 5 + (2 - \frac{\pi(2)^2}{4}) = 7 - \pi \approx 3.86

Compare values

The values are

f(-2)=3, f(-1)=4, f(2)=1, f(6)=5, f(8)=7-\pi

. The absolute minimum is 1.

AP Scoring Note — P7 + P8 + P9

P7 (Critical Points): Earned for considering

f'(x) = 0

.

P8 (Justification): Must show the values or a logical elimination of endpoints/points.

P9 (Answer): The minimum value is 1. (Note: AP asks for the *value*, not just the

x

coordinate).

Timed Practice

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Common Pitfalls

⚠
The 'Going Backwards' Integration ErrorWhen calculating $f(0)$ given $f(3) = 5$ , you need $5 + \int_3^0 f'(x) dx$ . Since the limits are upper-to-lower ( $3 > 0$ ), you must subtract the area under the graph. Students often forget to flip the sign.
⚠
Misidentifying Points of Inflection (POI)A POI occurs where $f''(x)$ changes sign, which means where the graph of $f'(x)$ changes from increasing to decreasing (a relative extremum on the $f'$ graph). Simply $f''(x)=0$ is not enough.
⚠
Vague Justifications (The 'It' Trap)Avoid using the word 'it' (e.g., 'It is increasing'). AP readers will not give credit. Specify the function: ' $f'(x)$ is positive,' or 'The slope of $f'$ is negative.'
⚠
Forgetting the Initial Value in FTCJust like in Particle Motion, finding $f(5)$ is not just the area from 0 to 5. It is $f(0) + \text{Area}$ . If you forget $f(0)$ , you lose the 'Answer' point.
⚠
Confusing $f'$ 's Value with $f'$ 's SlopeRemember: The y-value of the graph is $f'(x)$ (determines increase/decrease). The slope of the graph is $f''(x)$ (determines concavity).

Past AP Exams

Real AP exam questions on this topic

Official AP Calculus AB free-response questions, each with a full step-by-step solution.

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Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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On this page

Identify the 'Home Base'

Map out f(x)f(x)f(x)'s Behavior

Check the Endpoints for Absolute Extrema

Justify with Calculus Terms

Practice Questions

Apply FTC Part 1

Identify f(8) from the graph

AP Scoring Note — P1 + P2

Identify POI criteria

Locate points on graph

AP Scoring Note — P3 + P4

Calculate g(12)

Calculate g(0)

AP Scoring Note — P5 + P6

Identify Candidates

Evaluate g(x) at each candidate

Compare and Conclude

AP Scoring Note — P7 + P8 + P9

Calculate g(-6)

Calculate g(4)

Calculate g(6)

AP Scoring Note — P1, P2, P3

Identify Relationship

Solve for Critical Points

AP Scoring Note — P4, P5

Calculate h(6)

Calculate h'(6)

Calculate h''(6)

AP Scoring Note — P6, P7, P8, P9

Analyze the sign of f'(x)

Determine Extrema

AP Scoring Note — P1

Identify where f' is decreasing

State the reason

AP Scoring Note — P2 + P3

Test for Indeterminate Form

Apply L'Hospital's Rule

Evaluate the Limit

AP Scoring Note — P4 + P5 + P6

Identify Critical Points

Evaluate candidates using FTC

Compare values

AP Scoring Note — P7 + P8 + P9

Ready to challenge yourself?

Real AP exam questions on this topic

Gary Chang

Map out $f(x)$ 's Behavior