2023 AP Calculus AB FRQ Question 5: Tabular Data, Chain Rule & FTC

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2023.

Question 5

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Tabular Data, Chain Rule & FTC

Limits, Chain Rule & FTC

Hard

The functions

f

and

g

are twice differentiable. The table shown gives values of the functions and their first derivatives at selected values of

x

Values of f, f', g, and g'

x	0	2	4	7
f(x)	10	7	4	5
f′(x)	3/2	−8	3	6
g(x)	1	2	−3	0
g′(x)	5	4	2	8

Part AMedium2 points

Let

h

be the function defined by

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

. Find

h'(7)

. Show the work that leads to your answer.

Answer

h'(7)=12

Full Solution & Work

Apply the Chain Rule

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

Evaluate at x = 7

h'(7) = f'(g(7))\cdot g'(7) = f'(0)\cdot 8 = \frac{3}{2}\cdot 8 = 12

AP Scoring — 2 Points

**P1**: Either

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

h'(7)=f'(g(7))\cdot g'(7)

.
**P2**: Requires P1 — only for the answer 12 (if P1 is not earned, this point can still be earned only for the response

f'(0)\cdot 8=12

\frac{3}{2}\cdot 8

Part BHard3 points

Let

k

be a differentiable function such that

k'(x)=(f(x))^2\cdot g(x)

. Is the graph of

k

concave up or concave down at the point where

x=4

? Give a reason for your answer.

Answer

Concave down, since

k''(4)=-40<0

Full Solution & Work

Differentiate k'(x) using Product and Chain Rules

k''(x) = 2f(x)\cdot f'(x)\cdot g(x) + (f(x))^2 \cdot g'(x)

Evaluate at x = 4

k''(4) = 2(4)(3)(-3) + (4)^2(2) = -72+32=-40

Conclude

The graph of

k

is **concave down** at

x=4

because

k''(4)<0

(and

k''

is continuous, since

f

g

are twice differentiable).

AP Scoring — 3 Points

**P1**: Either

k''(x)=2f(x)\cdot f'(x)\cdot g(x)+(f(x))^2\cdot g'(x)

k''(4)=2f(4)\cdot f'(4)\cdot g(4)+(f(4))^2\cdot g'(4)

— several partially-correct expressions with a single product/chain-rule error still earn this point.
**P2**: Requires correctly computing

k''(4)=-40

with supporting work.
**P3**: A correct answer and reason consistent with the declared value of

k''(4)

Part CMedium1 point

Let

m

be the function defined by

m(x) = 5x^3+\displaystyle\int_0^x f'(t)\,dt

. Find

m(2)

. Show the work that leads to your answer.

Answer

m(2)=37

Full Solution & Work

Apply the Fundamental Theorem of Calculus

m(2) = 5(2)^3+\int_0^2 f'(t)\,dt = 40+\big(f(2)-f(0)\big) = 40+(7-10)=37

AP Scoring — 1 Point

**P1**: Earned only for the answer 37 (or equivalent) with supporting work equivalent to

5\cdot 8+(f(2)-f(0))

40+(7-10)

, or similar. An unsupported answer of just "37" does not earn this point.

Part DMedium3 points

Is the function

m

defined in part (c) increasing, decreasing, or neither at

x=2

? Justify your answer.

Answer

Increasing, because

m'(2)=52>0

Full Solution & Work

Differentiate m(x)

m'(x) = 15x^2+f'(x)

Evaluate at x = 2

m'(2) = 15(4)+f'(2)=60+(-8)=52

Conclude

The graph of

m

is **increasing** at

x=2

because

m'(2)=52>0

AP Scoring — 3 Points

**P1**: Considering

m'(x)

m'(2)

, or

m'

— this consideration may appear inside a justification statement.
**P2**: Correctly evaluates

m'(2)=15\cdot2^2+f'(2)

=60+f'(2)

, or

=60-8

— an unsupported response of

m'(2)=52

alone does not earn this point.
**P3**: An answer and justification consistent with the declared value of

m'(2)

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