2023
AP
®
Calculus AB
Sample Student Responses
and Scoring Commentary
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Inside:
Free-Response Question 5
Scoring Guidelines
Student Samples
Scoring Commentary
AP® Calculus AB/BC 2023 Scoring Guidelines
© 2023 College Board
Part B (AB): Graphing calculator not allowed
Question 5 9 points
General Scoring Notes
The model solution is presented using standard mathematical notation.
Answers (numeric or algebraic) need not be simplified. Answers given as a decimal approximation should be
correct to three places after the decimal point. Within each individual free-response question, at most one
point is not earned for inappropriate rounding.
(
)
( )
( )
(
)
0247
10 7 4 5
3
83 6
2
1 2 30
5428
x
fx
fx
gx
gx
′
′
−
−
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
Model Solution Scoring
(a)
Let
h
be the function defined by
( ) ( )
( )
.hx f gx=
Find
Show the work that leads to your answer.
( ) (
)
( )
( )
h x f gx g x
′
⋅
′′
=
( ) ( )
( )
( )
7 77h fg g
′
= ⋅
′′
Chain rule 1 point
( )
3
8 8 12
2
0f= ⋅= =
′
⋅
Answer 1 point
Scoring notes:
• The first point is earned for either
( ) ( )
( )
( )
h x f gx g x
′
⋅
′′
=
or
( ) (
)
( )
( )
7 7 7.h fg g
′
⋅
′′
=
• If the first point is earned, the second point is earned only for an answer of
12
(or equivalent).
• If the first point is not earned, the second point can be earned only for a response of either
( )
820 1f
′
⋅=
or
3
8.
2
⋅
• A response of
12
with no supporting work does not earn either point.
Total for part (a) 2 points
(b)
Let
k
be a differentiable function such that
( ) ( )
( )
( )
2
.k x f x gx
′
= ⋅
Is the graph of
k
concave up or
concave down at the point where
4?x =
Give a reason for your answer.
( ) ( )
( ) ( ) ( )
( )
( )
2
2k x fx f x gx fx gx
′′ ′ ′
= ⋅⋅+⋅
Product or chain rule 1 point
AP® Calculus AB/BC 2023 Scoring Guidelines
© 2023 College Board
( ) ( ) (
) (
) ( )
( )
( )
2
4 24 4 4 4 4k ff g f g
′′ ′ ′
= +⋅⋅ ⋅
( )
2
2 4 3 3 4 2 72 32 40⋅⋅⋅ ⋅= −+ =− + =−
1 point
The graph of
k
is concave down at the point where
4
x =
because
( )
40k
′′
<
and
k
′′
is continuous.
Answer with reason 1 point
Scoring notes:
• The first point is earned for either
( ) ( ) ( ) ( ) ( )
( )
( )
2
2k x fx f x gx fx gx
′′ ′ ′
= ⋅⋅+⋅
or
( ) ( ) ( ) ( ) ( )
( )
( )
2
4 2 4 4 4 4 4.k ffg f g⋅⋅
′
+ ⋅
′′ ′
=
• The first point is also earned by any of the following incorrect expressions, each of which has a
single error in the application of the product rule or the chain rule:
o
( ) ( ) ( )
( )
( )
2
2 fx gx fx gx+⋅⋅
′
or
( ) ( ) ( )
( )
( )
2
24 4 4 4fg f g
+⋅⋅
′
o
( ) ( ) ( )
( )
( )
2
2 f x gx f x g x⋅⋅
′′
+
or
( ) ( ) ( )
( )
( )
2
24 4 4 4fg f g
′
⋅
′
⋅ +
o
(
) ( )
( )
( )
( )
2
f x gx f x g x
+ ⋅
′
⋅
′
or
( )
( )
(
)
(
)
( )
2
44 4 4
fg f g
+⋅⋅
′′
o
( ) ( ) ( )
2 fx f x gx
′
⋅⋅
′
or
( ) ( ) ( )
24 4 4
ffg
′
⋅⋅
′
o Note: A response that presents one of these expressions cannot earn the second point.
• To earn the second point a response must correctly find
( )
4 40k
′′
= −
(or equivalent) with supporting
work.
• The third point is earned for an answer and reason that are consistent with any declared nonzero
value of
( )
4.k
′′
Total for part (b) 3 points
(c)
Let
m
be the function defined by
( ) ( )
3
0
5.
x
m x x f t dt
′
= +
∫
Find
Show the work that leads to
your answer.
( ) ( )
( )
( )
(
)
( )
2
0
2 5 40 2 0
40 7 10 37
8m f t dt f f
′
=+=+
= +−
⋅ −
=
∫
Answer with
supporting work
1 point
Scoring notes:
• The point is earned only for an answer of
37
(or equivalent) with supporting work equivalent to
( ) ( )
( )
528 0,ff+⋅ −
( ) ( )
( )
40 2 0 ,ff+−
( )
578
10 ,+⋅ −
or
( )
40 7 10 .
+−
• An answer of
37
with no supporting work does not earn the point.
Total for part (c) 1 point
(d) Is the function
m
defined in part (c) increasing, decreasing, or neither at
2?x =
Justify your answer.
( ) ( )
2
15mx x f x
′′
= +
Considers
( )
mx
′
1 point
( ) ( ) (
)
2 15 2 60 84 52mf
′′
=+=⋅ +− =
with
supporting work
1 point
AP® Calculus AB/BC 2023 Scoring Guidelines
© 2023 College Board
The graph of
m
is increasing at
2x =
because
( )
2 0.m
′
>
Answer with
justification
1 point
Scoring notes:
• The first point is earned for considering
( )
,
mx
′
( )
2,m
′
or
.
m
′
This consideration may appear in a
justification statement.
• The second point is earned for
( ) ( )
2
2 15 2 2 ,mf
′′
=⋅+
( ) ( )
2 60 2 ,mf
′′
= +
or
(
)
60
2
8m
′
=
−
but
is not earned for an unsupported response of
( )
2 52.m′=
• The third point is earned for an answer and justification consistent with any declared value of
(
)
2.m
′
Total for part (d) 3 points
Total for question 5 9 points
1 of 2
Sample 5A
2 of 2
Sample 5A
1 of 2
Sample 5B
2 of 2
Sample 5B
1 of 2
Sample 5C
2 of 2
Sample 5C
AP
®
Calculus AB 2023 Scoring Commentary
© 2023 College Board.
Visit College Board on the web: collegeboard.org.
Question 5
Note: Student samples are quoted verbatim and may contain spelling and grammatical errors.
Overview
In this problem students were given a table of selected values of the twice-differentiable functions
f
and
g
and of
their first derivatives.
In part (a) students are asked to find
( )
7h
′
for the function
( ) ( )
( )
.hx f gx=
A correct response will use the chain
rule to find
( ) ( )
( )
( )
,h x f gx g x
′
= ⋅
′′
then pull the appropriate values from the given table to find
(
)
7 12.
h
′
=
In part (b) students were told that
k
is a differentiable function such that
( ) ( )
( )
( )
2
k x f x gx⋅
′
=
and were asked
whether
k
is concave up or concave down at the point where
4.x =
A correct response will use the product and
chain rules to find
( )
kx
′′
and then evaluate
( )
4 40k
′′
= −
in order to determine that
k
is concave down at this
point.
In part (c) the function
( ) ( )
3
0
5
x
m x x f t dt
′
= +
∫
is defined and students were asked to find
( )
2.m
A correct
response will use the Fundamental Theorem of Calculus to find
( ) ( ) ( )
2
0
2 0,f t dt f f
′
= −
∫
then use the given table
to find
( )
2f
and
( )
0.f
Finally, a correct response will combine the difference of these values with
3
52⋅
to obtain
( )
2 37.m =
In part (d) students were asked whether this function
m
is increasing, decreasing, or neither at
2x =
and to provide
a justification for their answer. A correct response will use the Fundamental Theorem of Calculus to find
( ) (
)
2
2 15 2 2 52mf⋅
′′
= +=
and realize that, because
( )
2
m
′
is positive, the function must be increasing in a
neighborhood around
2.x =
Sample: 5A
Score: 9
The response earned 9 points: 2 points in part (a), 3 points in part (b), 1 point in part (c), and 3 points in part (d).
In part (a) the response earned the first point in line 1 on the right side with the correct chain rule. The numerical
expression
3
8
2
⋅
in line 2 on the right would have earned the second point with no simplification. In this case,
correct simplification earned the point with
(
)
7 12.h
′
=
In part (b) the response earned the first point in line 2 for the correct expression for
( )
.kx
′′
The expression
( )(
)( ) ( )
2
23 4 3 24−+
in line 4 would have earned the second point with no simplification. In this case, correct
simplification to
40−
in line 5 earned the point. The response earned the third point in line 6 for the correct answer
and reason, “concave down at
4
x =
because
( )
0.kx
′′
<
” This statement can be interpreted as
(
)
40k
′′
<
because
4x
=
was stated in the stem of the question.
In part (c) the numerical expression
( )
40 7 10+−
in line 3 would have earned the point with no simplification. In
this case, correct simplification to
37
in line 3 earned the point.
AP
®
Calculus AB 2023 Scoring Commentary
© 2023 College Board.
Visit College Board on the web: collegeboard.org.
Question 5 (continued)
In part (d) the response earned the first point in line 1 for considering
( )
.mx
′
The expression
( ) ( )
2
15 2 2f
′
+
would
have earned the second point with no simplification. In this case, correct simplification to
52
in line 2 earned the
point. The response earned the third point in line 4 and line 5 for the correct conclusion with the correct reasoning.
Sample: 5B
Score: 4
The response earned 4 points: 2 points in part (a), 1 point in part (b), 1 point in part (c), and no points in part (d).
In part (a) the response earned the first point in line 2 with the correct chain rule. The second point could have been
earned for the numeric expression
3
8
2
⋅
in line 4 but was simplified and the point was earned for the boxed answer
of
12.
In part (b) the expression for
(
)
4k
′′
in the form of a product rule with no evidence of a chain rule in line 2 earned
the first point but is not eligible for the second point. The response did not earn the third point because the reason
given “becaus the second derivative is pos but the first is neg” implies that concavity is based on both the first and
second derivatives.
In part (c) the point could have been earned for the numeric expression
40 3
+−
on the right but was simplified and
the point was earned for the boxed answer
( )
2 37m =
with supporting work.
In part (d) no points were earned because the response never considers
( )
mx
′
or
( )
2.m
′
Sample: 5C
Score: 3
The response earned 3 points: no points in part (a), no points in part (b), 1 point in part (c), and 2 points in part (d).
In part (a) the response did not earn the first point because the expression presented in line 1 on the right
( ) (
)
( )
(
)
h x f gx g x
′′ ′
= +
is not the correct chain rule. The second point was not earned because the boxed answer
( )
19
7
2
h
′
=
is incorrect.
In part (b) the response earned no points because no expression for
( )
kx
′′
or
( )
4k
′′
is present.
In part (c) the point could have been earned for the numeric expression
( )
40 3+−
in line 4 but was simplified and
the point was earned for the boxed answer
( )
2 37m =
with supporting work.
In part (d) the first point was earned in line 1 for considering
( )
mx
′
even though the expression presented for
( )
mx
′
is incorrect. The response did not earn the second point because the declared value
(
)
123
2
2
m
′
=
is incorrect. The
response earned the third point in lines 4 and 5 for the conclusion and justification consistent with the incorrect value
declared for
( )
2.m
′