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2021 AP Calculus AB FRQ Question 4: Graph Analysis with an Accumulation Function

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2021.

Question 4

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Graph Analysis with an Accumulation Function

Graph Analysis & Curve Sketching

Hard
Let ff be a continuous function defined on the closed interval −4≤x≤6-4\leq x\leq6. The graph of ff, consisting of four line segments connecting (−4,−2)(-4,-2), (−2,6)(-2,6), (0,4)(0,4), (2,−4)(2,-4), and (6,2)(6,2), is shown. Let GG be the function defined by G(x)=∫0xf(t) dtG(x)=\displaystyle\int_0^x f(t)\,dt.
2021 AP Calculus AB FRQ Question 4 Graph of f
Graph of $f$ on $[-4,6]$.
Part AMedium2 points
On what open intervals is the graph of GG concave up? Give a reason for your answer.

Answer

(−4,−2)(-4,-2) and (2,6)(2,6)
Full Solution & Work

Relate concavity of G to the slope of f

G′(x)=f(x)G'(x)=f(x), so G′′(x)=f′(x)G''(x)=f'(x) equals the (constant) slope of ff on each line segment. GG is concave up where f′(x)>0f'(x)>0, i.e. where ff has positive slope.

Compute each segment's slope

(−4,−2)→(−2,6)(-4,-2)\to(-2,6): slope =82=4>0=\frac{8}{2}=4>0.
(−2,6)→(0,4)(-2,6)\to(0,4): slope =−22=−1<0=\frac{-2}{2}=-1<0.
(0,4)→(2,−4)(0,4)\to(2,-4): slope =−82=−4<0=\frac{-8}{2}=-4<0.
(2,−4)→(6,2)(2,-4)\to(6,2): slope =64=1.5>0=\frac{6}{4}=1.5>0.

Conclude

GG is concave up where the slope of ff is positive: on **(−4,−2)(-4,-2) and (2,6)(2,6)**.

AP Scoring — 2 Points

**P1**: Correctly relates G′′=f′G''=f' to the slopes of the line segments.
**P2**: Correct intervals, (−4,−2)(-4,-2) and (2,6)(2,6).
Part BHard3 points
Let PP be the function defined by P(x)=G(x)⋅f(x)P(x)=G(x)\cdot f(x). Find P′(3)P'(3).

Answer

P′(3)=1.375P'(3)=1.375
Full Solution & Work

Apply the Product Rule

P′(x)=G′(x)f(x)+G(x)f′(x)=(f(x))2+G(x)f′(x)P'(x) = G'(x)f(x)+G(x)f'(x) = \big(f(x)\big)^2+G(x)f'(x)

Find f(3) and f'(3)

On [2,6][2,6], f(x)=−4+1.5(x−2)f(x)=-4+1.5(x-2). So f(3)=−4+1.5=−2.5f(3)=-4+1.5=-2.5, and f′(3)=1.5f'(3)=1.5 (the segment's constant slope).

Find G(3)

G(3)=∫03f(t) dt=∫02f(t) dt+∫23f(t) dt=4+(−4)2(2)+−4+(−2.5)2(1)=0−3.25=−3.25G(3)=\int_0^3 f(t)\,dt = \int_0^2 f(t)\,dt+\int_2^3 f(t)\,dt = \frac{4+(-4)}{2}(2)+\frac{-4+(-2.5)}{2}(1) = 0-3.25=-3.25

Combine

P′(3)=(f(3))2+G(3)f′(3)=(−2.5)2+(−3.25)(1.5)=6.25−4.875=1.375P'(3) = \big(f(3)\big)^2+G(3)f'(3) = (-2.5)^2+(-3.25)(1.5) = 6.25-4.875=1.375

AP Scoring — 3 Points

**P1**: Correct Product Rule setup, P′(x)=(f(x))2+G(x)f′(x)P'(x)=(f(x))^2+G(x)f'(x).
**P2**: Correctly finds f(3)=−2.5f(3)=-2.5, f′(3)=1.5f'(3)=1.5, and G(3)=−3.25G(3)=-3.25.
**P3**: Correct final answer, 1.375.
Part CHard2 points
Find lim⁡x→2G(x)x2−2x\lim\limits_{x\to2} \dfrac{G(x)}{x^2-2x}.

Answer

−2-2
Full Solution & Work

Confirm the indeterminate form

G(2)=∫02f(t) dt=0G(2)=\int_0^2 f(t)\,dt=0 (computed in part (b)), and x2−2x=x(x−2)→0x^2-2x=x(x-2)\to0 as x→2x\to2. This is 00\frac00, so L'Hôpital's Rule applies.

Apply L'Hôpital's Rule

lim⁡x→2G(x)x2−2x=lim⁡x→2G′(x)2x−2=lim⁡x→2f(x)2x−2\lim_{x\to2}\frac{G(x)}{x^2-2x} = \lim_{x\to2}\frac{G'(x)}{2x-2} = \lim_{x\to2}\frac{f(x)}{2x-2}

Evaluate

f(2)=−4f(2)=-4 (continuous at the shared endpoint of the two segments), and 2(2)−2=22(2)-2=2:
=−42=−2= \frac{-4}{2}=-2

AP Scoring — 2 Points

**P1**: Confirms the 00\frac00 form and applies L'Hôpital's Rule (or an equivalent factoring approach).
**P2**: Correct final answer, −2-2.
Part DMedium2 points
Find the average rate of change of GG on the interval [−4,2][-4,2]. Does the Mean Value Theorem guarantee a value cc, −4<c<2-4<c<2, for which G′(c)G'(c) is equal to this average rate of change? Justify your answer.

Answer

Average rate of change =73=\dfrac{7}{3}; yes, the MVT applies.
Full Solution & Work

Find G(-4) and G(2)

G(2)=0G(2)=0 (from part b). G(−4)=∫0−4f(t) dt=−∫−40f(t) dtG(-4)=\int_0^{-4}f(t)\,dt = -\int_{-4}^0 f(t)\,dt, and ∫−40f(t) dt=−2+62(2)+6+42(2)=4+10=14\int_{-4}^0 f(t)\,dt = \frac{-2+6}{2}(2)+\frac{6+4}{2}(2) = 4+10=14, so G(−4)=−14G(-4)=-14.

Compute the average rate of change

G(2)−G(−4)2−(−4)=0−(−14)6=146=73\frac{G(2)-G(-4)}{2-(-4)} = \frac{0-(-14)}{6} = \frac{14}{6}=\frac73

Check the hypotheses of the MVT

G(x)=∫0xf(t) dtG(x)=\int_0^x f(t)\,dt with ff continuous everywhere on [−4,6][-4,6], so GG is differentiable (with G′=fG'=f) — and hence continuous — on all of [−4,2][-4,2].

Conclude

Since GG is differentiable on (−4,2)(-4,2) and continuous on [−4,2][-4,2], the **Mean Value Theorem** guarantees a c∈(−4,2)c\in(-4,2) with G′(c)=73G'(c)=\frac73.

AP Scoring — 2 Points

**P1**: Correctly computes the average rate of change, 73\frac73.
**P2**: Correct "yes" with justification citing differentiability of GG (via continuity of ff).

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