Calculus-Master
Start Free Trial

Free Response · Interpreting the derivative to understand the original function

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

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

On this page

Key Formulashover to see usage
  • The Fundamental Theorem of Calculus (Accumulation)f(x)=f(a)+∫axf′(t) dtf(x) = f(a) + \int_a^x f'(t)\,dt
    When to useTo find a specific value f(x)f(x), start at the known point f(a)f(a) and add the 'net area' from aa to xx.
  • Area as Net Change∫abf′(x) dx=f(b)−f(a)\int_a^b f'(x)\,dx = f(b) - f(a)
    When to useThe definite integral is the area between the f′(x)f'(x) graph and the x-axis. Areas above the axis are positive; areas below are negative.
  • First Derivative Test (Extrema)f′(x)=0 or undefinedf'(x) = 0 \text{ or undefined}
    When to useCritical points occur where the graph of f′f' crosses or touches the x-axis. f′f' changes ++ to −- (Relative Max); f′f' changes −- to ++ (Relative Min).
  • Concavity and f′′(x)f''(x)f′′(x)=ddx[f′(x)]f''(x) = \frac{d}{dx}[f'(x)]
    When to useThe slope of the f′(x)f'(x) graph is f′′(x)f''(x). If f′(x)f'(x) is increasing (positive slope), f(x)f(x) is concave up.
Step-by-Step SOP
  1. 1

    Identify the 'Home Base'

    Locate the given point (a,f(a))(a, f(a)). This is your starting constant for all FTC calculations. If you go backwards (e.g., ∫52\int_5^2), remember to flip the sign of the area.
  2. 2

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

    Scan the f′f' graph: Where is it above the x-axis (ff increases)? Where is it below (ff decreases)? Where are the 'peaks and valleys' of f′f' (Inflection points of ff)?
  3. 3

    Check the Endpoints for Absolute Extrema

    If asked for the 'Absolute Maximum/Minimum' on [a,b][a, b], you MUST test the critical points AND the endpoints aa and bb. This is the Candidates Test.
  4. 4

    Justify with Calculus Terms

    Never say 'the graph goes up' or 'it changes signs.' Always say 'f′(x)f'(x) changes from positive to negative' or 'f′(x)f'(x) is increasing.'

Practice Questions

123

Question 1

2025 AP Calculus AB Exam — FRQ 4

Medium
The continuous function ff is defined on the closed interval −6≤x≤12-6 \le x \le 12. The graph of ff, consisting of two semicircles and one line segment, is shown in the figure. Let gg be the function defined by g(x)=∫6xf(t) dtg(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)g'(8). Give a reason for your answer.
Need a Hint?
By the Fundamental Theorem of Calculus, if g(x)=∫6xf(t) dtg(x) = \int_{6}^{x} f(t) \, dt, then g′(x)=f(x)g'(x) = f(x). Find the value of f(8)f(8) from the graph.
Show Solution

Apply FTC Part 1

g′(x)=ddx∫6xf(t) dt=f(x)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)(6, 0) and (12,3)(12, 3). At x=8x = 8, f(8)=1f(8) = 1.

AP Scoring Note — P1 + P2

P1 (Reason): Earned for explicitly stating g′(x)=f(x)g'(x) = f(x) or g′(8)=f(8)g'(8) = f(8).

P2 (Answer): g′(8)=1g'(8) = 1. Note: Simply writing g′(8)=1g'(8) = 1 without mentioning f(8)f(8) might lose the reasoning point depending on the year's stringency.
BPart B
Medium
Find all values of xx in the open interval −6<x<12-6 < x < 12 at which the graph of gg has a point of inflection. Give a reason for your answer.
Need a Hint?
A point of inflection for gg occurs when g′′(x)g''(x) changes sign. Since g′(x)=f(x)g'(x) = f(x), g′′(x)=f′(x)g''(x) = f'(x). Look for where f(x)f(x) changes from increasing to decreasing or vice versa.
Show Solution

Identify POI criteria

Inflection points of gg occur where g′(x)=f(x)g'(x) = f(x) changes from increasing to decreasing (relative extrema of ff).

Locate points on graph

ff changes direction at x=−3x = -3 (dec to inc), x=3x = 3 (inc to dec), and x=6x = 6 (dec to inc).

AP Scoring Note — P3 + P4

P3 (Answer): Must list all three: x=−3,3,6x = -3, 3, 6.

P4 (Reason): You must tie the reason to the graph of ff. Correct: 'ff changes from increasing to decreasing or vice versa'. Incorrect: 'The slope of the graph changes' (too vague).
CPart C
Medium
Find g(12)g(12) and g(0)g(0). Label your answers.
Need a Hint?
Use geometry to find the area under the curve. Remember g(0)=∫60f(t) dt=−∫06f(t) dtg(0) = \int_{6}^{0} f(t) \, dt = -\int_{0}^{6} f(t) \, dt.
Show Solution

Calculate g(12)

g(12)=∫612f(t) dt=12(base)(height)=12(6)(3)=9g(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)=∫60f(t) dt=−∫06f(t) dtg(0) = \int_{6}^{0} f(t) \, dt = -\int_{0}^{6} f(t) \, dt. The area from 0 to 6 is a semicircle with r=3r=3: 12π(3)2=9π2\frac{1}{2}\pi(3)^2 = \frac{9\pi}{2}. Therefore, g(0)=−9π2g(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 12(6)(3)\frac{1}{2}(6)(3) to 9 to get the point.
DPart D
Hard
Find the value of xx at which gg attains an absolute minimum on [−6,12][-6, 12]. Justify your answer.
Need a Hint?
Use the Candidates Test. Check the endpoints (x=−6,12x = -6, 12) and critical points where g′(x)=f(x)=0g'(x) = f(x) = 0.
Show Solution

Identify Candidates

Critical points (f(x)=0f(x)=0): x=0x = 0 and x=6x = 6. Endpoints: x=−6x = -6 and x=12x = 12.

Evaluate g(x) at each candidate

g(−6)=∫6−6f(t) dt=−(Area1+Area2)=−(9π2−9π2)=0g(-6) = \int_{6}^{-6} f(t) \, dt = -(\text{Area}_1 + \text{Area}_2) = -(\frac{9\pi}{2} - \frac{9\pi}{2}) = 0.
g(0)=−9π2≈−14.137g(0) = -\frac{9\pi}{2} \approx -14.137.
g(6)=0g(6) = 0.
g(12)=9g(12) = 9.

Compare and Conclude

The smallest value is −9π2-\frac{9\pi}{2} at x=0x = 0.

AP Scoring Note — P7 + P8 + P9

P7 (Critical Points): Earned for considering where f(x)=0f(x) = 0.

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

P9 (Answer): x=0x = 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 ff, shown for −6≤x≤7-6 \leq x \leq 7, has a horizontal tangent at x=−2x = -2 and is linear for 0≤x≤70 \le x \le 7. Let RR be the region in the second quadrant bounded by the graph of ff, the vertical line x=−6x = -6, and the xx- and yy-axes. Region RR 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 gg is defined by g(x)=∫0xf(t) dtg(x) = \int_{0}^{x} f(t) \, dt. Find the values of g(−6)g(-6), g(4)g(4), and g(6)g(6).
Need a Hint?
Remember that ∫abf(t)dt=−∫baf(t)dt\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)=∫0−6f(t) dt=−∫−60f(t) dt=−(Area of R)=−12g(-6) = \int_{0}^{-6} f(t) \, dt = -\int_{-6}^{0} f(t) \, dt = -(\text{Area of } R) = -12.

Calculate g(4)

g(4)=∫04f(t) dtg(4) = \int_{0}^{4} f(t) \, dt. This is a triangle with base 4 and height 2. Area = 12(4)(2)=4\frac{1}{2}(4)(2) = 4.

Calculate g(6)

g(6)=g(4)+∫46f(t) dt=4+12(2)(−1)=4−1=3g(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 gg defined in part (a), find all values of xx in the interval 0≤x≤60 \le x \le 6 at which the graph of gg has a critical point. Give a reason for your answer.
Need a Hint?
A critical point of gg occurs where g′(x)=0g'(x) = 0 or is undefined. Use FTC to relate g′g' to ff.
Show Solution

Identify Relationship

By FTC, g′(x)=f(x)g'(x) = f(x).

Solve for Critical Points

g′(x)=f(x)=0  ⟹  x=4g'(x) = f(x) = 0 \implies x = 4 (within the interval [0,6][0, 6]).

AP Scoring Note — P4, P5

P4 (FTC): Earned for explicitly writing g′(x)=f(x)g'(x) = f(x).

P5 (Answer+Reason): x=4x = 4 with the reason that g′(4)=0g'(4) = 0. Reporting extra points in the interval loses this point.
CPart C
Medium
The function hh is defined by h(x)=∫−6xf′(t) dth(x) = \int_{-6}^{x} f'(t) \, dt. Find the values of h(6)h(6), h′(6)h'(6), and h′′(6)h''(6). Show the work that leads to your answers.
Need a Hint?
For h(6)h(6), use FTC: ∫abf′(t)dt=f(b)−f(a)\int_a^b f'(t)dt = f(b) - f(a). For h′h' and h′′h'', differentiate h(x)h(x) and use the properties of the graph of f.
Show Solution

Calculate h(6)

h(6)=∫−66f′(t) dt=f(6)−f(−6)=−1−0.5=−1.5h(6) = \int_{-6}^{6} f'(t) \, dt = f(6) - f(-6) = -1 - 0.5 = -1.5.

Calculate h'(6)

h′(x)=ddx∫−6xf′(t) dt=f′(x)h'(x) = \frac{d}{dx} \int_{-6}^{x} f'(t) \, dt = f'(x). Since ff is linear on [0,7][0, 7], f′(6)f'(6) is the slope of the line: −1−26−0=−36=−12\frac{-1 - 2}{6 - 0} = -\frac{3}{6} = -\frac{1}{2}. So h′(6)=−0.5h'(6) = -0.5.

Calculate h''(6)

h′′(x)=f′′(x)h''(x) = f''(x). Since ff is linear on [0,7][0, 7], its second derivative is 00. So h′′(6)=0h''(6) = 0.

AP Scoring Note — P6, P7, P8, P9

P6 (FTC): Use of f(6)−f(−6)f(6)-f(-6).

P7 (h(6) Value): Correct calculation of −1.5-1.5.

P8 (h'(6)): Earned for stating h′(6)=f′(6)h'(6) = f'(6) and the value −1/2-1/2.

P9 (h''(6)): Earned for 00.

Question 3

2023 AP Calculus AB Exam — FRQ 4

Hard
The function ff is defined on the closed interval [−2,8][-2, 8] and satisfies f(2)=1f(2) = 1. The graph of f′f', the derivative of ff, 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 ff have a relative minimum, a relative maximum, or neither at x=6x = 6? Give a reason for your answer.
Need a Hint?
Relative extrema occur where f′(x)f'(x) changes sign. Check the graph of f′f' at x=6x=6.
Show Solution

Analyze the sign of f'(x)

From the graph, f′(x)>0f'(x) > 0 on the interval (2,6)(2, 6) and f′(x)>0f'(x) > 0 on the interval (6,8)(6, 8).

Determine Extrema

Since f′(x)f'(x) does not change sign at x=6x = 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′f' 'does not change sign'. Simply saying f′(6)=0f'(6)=0 is not enough.
BPart B
Medium
On what open intervals, if any, is the graph of ff concave down? Give a reason for your answer.
Need a Hint?
A function is concave down when f′′(x)<0f''(x) < 0, which means f′(x)f'(x) is decreasing.
Show Solution

Identify where f' is decreasing

Looking at the graph, f′(x)f'(x) is decreasing on the intervals (−2,0)(-2, 0) and (4,6)(4, 6).

State the reason

The graph of ff is concave down on (−2,0)(-2, 0) and (4,6)(4, 6) because f′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′f' (decreasing). Mentioning f′′<0f'' < 0 alone may not be enough if not tied to the given graph of f′f'.
CPart C
Medium
Find the value of lim⁡x→26f(x)−3xx2−5x+6\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 ff is differentiable, it is continuous. lim⁡x→2(6f(x)−3x)=6(1)−3(2)=0\lim_{x \to 2} (6f(x) - 3x) = 6(1) - 3(2) = 0.
lim⁡x→2(x2−5x+6)=22−5(2)+6=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→26f′(x)−32x−5\lim_{x \to 2} \frac{6f'(x) - 3}{2x - 5}.

Evaluate the Limit

From the graph, f′(2)=0f'(2) = 0.
Limit = 6(0)−32(2)−5=−3−1=3\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 ff on the closed interval [−2,8][-2, 8]. Justify your answer.
Need a Hint?
Use the Candidates Test: Check endpoints x=−2,8x = -2, 8 and critical points where f′(x)=0f'(x) = 0.
Show Solution

Identify Critical Points

f′(x)=0f'(x) = 0 at x=−1,2,6x = -1, 2, 6.

Evaluate candidates using FTC

f(2)=1f(2) = 1 (Given).
f(−2)=f(2)+∫2−2f′(t) dt=1−(net area from -2 to 2)=1−(−2)=3f(-2) = f(2) + \int_{2}^{-2} f'(t) \, dt = 1 - (\text{net area from -2 to 2}) = 1 - (-2) = 3.
f(6)=f(2)+∫26f′(t) dt=1+(area of triangle)=1+12(4)(2)=5f(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)+∫68f′(t) dt=5+(2−π(2)24)=7−π≈3.86f(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−πf(-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)=0f'(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 xx coordinate).

Timed Practice

Ready to challenge yourself?

Take a 20-minute timed mock exam to test your understanding.

Start Timed Test (20m)
Common Pitfalls
  • ⚠
    The 'Going Backwards' Integration ErrorWhen calculating f(0)f(0) given f(3)=5f(3) = 5, you need 5+∫30f′(x)dx5 + \int_3^0 f'(x) dx. Since the limits are upper-to-lower (3>03 > 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)f''(x) changes sign, which means where the graph of f′(x)f'(x) changes from increasing to decreasing (a relative extremum on the f′f' graph). Simply f′′(x)=0f''(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)f'(x) is positive,' or 'The slope of f′f' is negative.'
  • ⚠
    Forgetting the Initial Value in FTCJust like in Particle Motion, finding f(5)f(5) is not just the area from 0 to 5. It is f(0)+Areaf(0) + \text{Area}. If you forget f(0)f(0), you lose the 'Answer' point.
  • ⚠
    Confusing f′f''s Value with f′f''s SlopeRemember: The y-value of the graph is f′(x)f'(x) (determines increase/decrease). The slope of the graph is f′′(x)f''(x) (determines concavity).