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2026 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 2026.

Question 4

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Analyzing a Function from its Derivative Graph

Graph Analysis & Curve Sketching

Hard
Let ff be a twice-differentiable function on [−4,4][-4, 4] with f(2)=3f(2) = 3. The graph of f′f', the derivative of ff, is shown, passing through the points (−4,−0.5)(-4, -0.5), (−3,−1)(-3, -1), (1,3)(1, 3), (2,1.5)(2, 1.5), and (4,5)(4, 5).
2026 AP Calculus AB FRQ Question 4 Graph
Part AEasy2 points
For x>0x > 0, the function g is defined by g(x)=f(x)−ln⁡xg(x) = f(x) - \ln x. Find g′(2)g'(2). Show the work that leads to your answer.

Answer

g′(2)=f′(2)−12=1.5−0.5=1g'(2) = f'(2) - \dfrac{1}{2} = 1.5 - 0.5 = 1
Full Solution & Work

Differentiate g(x)

g′(x)=f′(x)−1xg'(x) = f'(x) - \frac{1}{x}

Evaluate at x = 2

From the graph of f′f': f′(2)=1.5f'(2) = 1.5.
g′(2)=f′(2)−12=1.5−0.5=1g'(2) = f'(2) - \frac{1}{2} = 1.5 - 0.5 = 1

AP Scoring — 2 Points

**P1**: g′(x)=f′(x)−1xg'(x) = f'(x) - \frac{1}{x} (correct differentiation).
**P2**: Correct answer g′(2)=1g'(2) = 1.
Part BMedium2 points
Find all values of xx on the open interval (0,3)(0, 3)at which the graph of ff has a point of inflection. Give a reason for your answer.

Answer

ff has a point of inflection at x=1x = 1 because f′′f'' changes sign there (from positive to negative on the graph of f′f').
Full Solution & Work

Recall: inflection points occur where f'' changes sign

Since f′′=(f′)′f'' = (f')', inflection points of ff occur where f′f' changes from increasing to decreasing (or vice versa), i.e., where f′f' has a local extremum.

Read the graph of f'

On (0,3)(0, 3): f′f' increases then decreases, with a local maximum at x=1x = 1. At x=1x = 1, f′f' changes from increasing to decreasing, so f′′f'' changes from ++ to −-.

Conclusion

There is exactly one point of inflection: x=1x = 1. At x=1x = 1, f′′f'' changes sign from positive to negative.

AP Scoring — 2 Points

**P1**: Identifies x=1x = 1.
**P2**: Justification — f′f' changes from increasing to decreasing at x=1x = 1, so f′′f'' changes sign.
Part CMedium2 points
On [−4,4][-4, 4], on what open intervals, if any, is the graph of ff both increasing and concave down? Give a reason for your answer.

Answer

ff is both increasing and concave down on the interval (1,3)(1, 3).
Full Solution & Work

Condition 1: Increasing ($f'(x) > 0$)

The graph of ff increases when its derivative f′f' is positive. Based on the provided graph, f′(x)>0f'(x) > 0 on the intervals (−2,3)(-2, 3) and (3,4)(3, 4).

Condition 2: Concave Down ($f''(x) < 0$)

The graph of ff is concave down when f′′<0f'' < 0, which corresponds to where f′f' is decreasing. From the graph, f′f' decreases on the intervals (−4,−3)(-4, -3) and (1,3)(1, 3).

Intersection of Conditions

To satisfy both conditions, we find the intersection of x∈(−2,3)∪(3,4)x \in (-2, 3) \cup (3, 4) and x∈(−4,−3)∪(1,3)x \in (-4, -3) \cup (1, 3). The only interval that satisfies both is (1,3)(1, 3).

AP Scoring — 2 Points

P1: Correct interval (1,3)(1, 3).
P2: Reason — States that ff is increasing because f′>0f' > 0 and concave down because f′f' is decreasing on this interval.
Part DHard4 points
On [−4,4][-4, 4], find the value of xx at which ff has an absolute minimum and the value of xx at which ff has an absolute maximum. Give reasons for your answers.

Answer

Absolute minimum at x=−2x = -2; absolute maximum at x=4x = 4.
Full Solution & Work

Identify Candidate Points

Absolute extrema can only occur at endpoints x=−4,4x = -4, 4 or critical points where f′(x)=0f'(x) = 0. From the graph, f′(x)=0f'(x) = 0 at x=−2x = -2 and x=3x = 3.

Analyze the Absolute Minimum

On the interval [−4,−2][-4, -2], f′(x)<0f'(x) < 0, so ff is decreasing. On [−2,4][-2, 4], f′(x)≥0f'(x) \ge 0 (except at x=3x=3 where it is 0), so ff is increasing. Since the function decreases from x=−4x = -4 to x=−2x = -2 and increases from x=−2x = -2 to x=4x = 4, the absolute minimum must occur at x=−2x = -2.

Analyze the Absolute Maximum

Since ff is increasing on the entire interval [−2,4][-2, 4], we only need to compare the endpoints f(−4)f(-4) and f(4)f(4). Because the area under f′f' from x=−2x = -2 to x=4x = 4 is much larger than the magnitude of the area from x=−4x = -4 to x=−2x = -2, the absolute maximum occurs at the right endpoint x=4x = 4.

AP Scoring — 4 Points

P1: Identifies x=−2x = -2 and x=3x = 3 as critical points.
P2: Justifies absolute minimum at x=−2x = -2 using the sign change of f′f'.
P3: Justifies absolute maximum at x=4x = 4 by noting ff increases on [−2,4][-2, 4].
P4: Correct values for both.

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