2026 AP Calculus AB FRQ Question 6: Tabular Data, Chain Rule & FTC

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2026.

Question 6

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Tabular Data, Chain Rule & FTC

Limits, Chain Rule & FTC

Hard

The function

f

is twice differentiable. The table gives values of

f

and

f'

at selected values of

x

Values of f and f'

x	0	2	3	6
f(x)	−1	3	8	5
f′(x)	−5	4	9	−2

Part AEasy2 points

Find

\lim_{x \to 2} \dfrac{f(x)}{x}

, or state that it does not exist.

Answer

\lim_{x \to 2} \dfrac{f(x)}{x} = \dfrac{f(2)}{2} = \dfrac{3}{2}

Full Solution & Work

Check if direct substitution applies

Since

f

is differentiable (hence continuous), we can substitute directly:

\lim_{x \to 2} \frac{f(x)}{x} = \frac{f(2)}{2} = \frac{3}{2}

AP Scoring — 2 Points

**P1**: Recognizes direct substitution is valid (no indeterminate form).
**P2**: Correct answer

\frac{3}{2}

Part BMedium3 points

Let

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

. Find

g'(2)

. Show your work.

Answer

g'(2) = f'(f(2)) \cdot f'(2) = f'(3) \cdot 4 = 9 \cdot 4 = 36

Full Solution & Work

Apply the Chain Rule

g'(x) = f'(f(x)) \cdot f'(x)

Evaluate at x = 2

g'(2) = f'(f(2)) \cdot f'(2) = f'(3) \cdot f'(2)

From the table:

f(2) = 3

f'(2) = 4

f'(3) = 9

g'(2) = 9 \cdot 4 = 36

AP Scoring — 3 Points

**P1**: Correct chain rule setup

f'(f(x)) \cdot f'(x)

.
**P2**: Substitutes

f(2) = 3

to get

f'(3)

.
**P3**: Correct answer 36.

Part CHard3 points

Let

h

be differentiable with

h(0) = 10

and

h'(x) = f'(3x)

. Find

h(2)

. Show your work.

Answer

h(2) = h(0) + \int_0^2 f'(3x)\,dx = 10 + \frac{1}{3}[f(6) - f(0)] = 10 + \frac{1}{3}(6) = 12

Full Solution & Work

Use FTC: h(2) = h(0) + integral

h(2) = h(0) + \int_0^2 h'(x)\,dx = 10 + \int_0^2 f'(3x)\,dx

U-substitution: u = 3x

Let

u = 3x

du = 3\,dx

, so

dx = \frac{du}{3}

. When

x=0

u=0

; when

x=2

u=6

\int_0^2 f'(3x)\,dx = \frac{1}{3}\int_0^6 f'(u)\,du = \frac{1}{3}[f(6) - f(0)]

Evaluate using table values

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

h(2) = 10 + 2 = 12

AP Scoring — 3 Points

**P1**: FTC setup

h(2) = 10 + \int_0^2 f'(3x)\,dx

.
**P2**: Correct

u

-substitution giving

\frac{1}{3}[f(6)-f(0)]

.
**P3**: Correct answer

h(2) = 12

Part D-iEasy1 point

Let

k(x) = \int_0^x t^2 f(t)\,dt

. Find

k'(x)

Answer

k'(x) = x^2 f(x)

Full Solution & Work

Apply FTC Part 1

By the Fundamental Theorem of Calculus Part 1:

k'(x) = x^2 f(x)

AP Scoring — 1 Point

**P1**: Correct answer

k'(x) = x^2 f(x)

Part D-iiHard3 points

Find

k''(3)

. Show your work.

Answer

k''(3) = 2(3)f(3) + (3)^2 f'(3) = 6(8) + 9(9) = 48 + 81 = 129

Full Solution & Work

Differentiate k'(x) using the Product Rule

k'(x) = x^2 f(x)

k''(x) = 2x\,f(x) + x^2 f'(x)

Evaluate at x = 3

From the table:

f(3) = 8

f'(3) = 9

k''(3) = 2(3)(8) + (3)^2(9) = 48 + 81 = 129

AP Scoring — 3 Points

**P1**: Correct product rule differentiation of

k'(x)

.
**P2**: Substitutes

f(3) = 8

and

f'(3) = 9

.
**P3**: Correct answer 129.

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