AP® CALCULUS BC
2011 SCORING GUIDELINES (Form B)
Question 4
The graph of the differentiable function y = f ( x ) with domain
0 ≤ x ≤ 10 is shown in the figure above. The area of the region enclosed
between the graph of f and the x-axis for 0 ≤ x ≤ 5 is 10, and the area of
the region enclosed between the graph of f and the x-axis for 5 ≤ x ≤ 10
is 27. The arc length for the portion of the graph of f between x = 0 and
x = 5 is 11, and the arc length for the portion of the graph
of f between x = 5 and x = 10 is 18. The function f has exactly two
critical points that are located at x = 3 and x = 8.
(a) Find the average value of f on the interval 0 ≤ x ≤ 5.
(b) Evaluate
10
∫0
( 3 f ( x ) + 2 ) dx. Show the computations that lead to
your answer.
(c) Let g ( x ) =
x
∫5
f ( t ) dt. On what intervals, if any, is the graph of g both concave up and decreasing? Explain
your reasoning.
(d) The function h is defined by h( x ) = 2 f
( 2x ). The derivative of h is h′( x ) = f ′( 2x ). Find the arc length of
the graph of y = h( x ) from x = 0 to x = 20.
(a) Average value =
(b)
10
∫0
1 5
−10
f ( x ) dx =
= −2
5 ∫0
5
( 3 f ( x ) + 2 ) dx = 3
(∫
5
0
f ( x ) dx +
10
∫5
1 : answer
)
f ( x ) dx + 20
2 : answer
= 3 ( −10 + 27 ) + 20 = 71
(c) g ′( x ) = f ( x )
g ′( x ) < 0 on 0 < x < 5
g ′( x ) is increasing on 3 < x < 8.
The graph of g is concave up and decreasing on 3 < x < 5.
(d) Arc length =
Let u =
20
⌠
⎮
⌡0
20
∫0
1 + ( h′( x ) )2 dx = ⌠
⎮
20
⌡0
( ( 2x ))
1+ f′
2
dx
x
1
. Then du = dx and
2
2
( ( ))
1+ f′
x
2
2
dx = 2∫
10
0
⎧ 1 : g ′( x ) = f ( x )
⎪
3 : ⎨ 1 : analysis
⎪⎩ 1 : answer and reason
⎧ 1 : integral
⎪
3 : ⎨ 1 : substitution
⎪⎩ 1 : answer
1 + ( f ′( u ) )2 du = 2 (11 + 18 ) = 58
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AP® CALCULUS BC
2011 SCORING GUIDELINES
Question 4
The continuous function f is defined on the interval − 4 ≤ x ≤ 3.
The graph of f consists of two quarter circles and one line
segment, as shown in the figure above.
Let g ( x ) = 2 x +
x
∫0 f ( t ) dt.
(a) Find g ( −3) . Find g ′( x ) and evaluate g ′( −3) .
(b) Determine the x-coordinate of the point at which g has an
absolute maximum on the interval − 4 ≤ x ≤ 3.
Justify your answer.
(c) Find all values of x on the interval − 4 < x < 3 for which
the graph of g has a point of inflection. Give a reason for
your answer.
(d) Find the average rate of change of f on the interval
− 4 ≤ x ≤ 3. There is no point c, − 4 < c < 3, for which f ′( c ) is equal to that average rate of change.
Explain why this statement does not contradict the Mean Value Theorem.
(a)
g ( −3) = 2 ( −3) +
−3
∫0
f ( t ) dt = −6 −
g ′( x ) = 2 + f ( x )
9π
4
⎧ 1 : g ( −3)
⎪
3 : ⎨ 1 : g ′( x )
⎪⎩ 1 : g ′( −3)
g ′( −3) = 2 + f ( −3) = 2
(b) g ′( x ) = 0 when f ( x ) = −2. This occurs at x =
5
.
2
5
5
and g ′( x ) < 0 for < x < 3.
2
2
5
Therefore g has an absolute maximum at x = .
2
g ′( x ) > 0 for − 4 < x <
(c)
g ′′( x ) = f ′( x ) changes sign only at x = 0. Thus the graph
of g has a point of inflection at x = 0.
(d) The average rate of change of f on the interval − 4 ≤ x ≤ 3 is
f ( 3) − f (− 4)
2
=− .
3 − (− 4)
7
To apply the Mean Value Theorem, f must be differentiable
at each point in the interval − 4 < x < 3. However, f is not
differentiable at x = −3 and x = 0.
⎧ 1 : considers g ′( x ) = 0
⎪
3 : ⎨ 1 : identifies interior candidate
⎪⎩ 1 : answer with justification
1 : answer with reason
2:
{
1 : average rate of change
1 : explanation
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AP® CALCULUS BC
2012 SCORING GUIDELINES
Question 3
Let f be the continuous function defined on [ − 4, 3]
whose graph, consisting of three line segments and a
semicircle centered at the origin, is given above. Let g
x
1
be the function given by g ( x ) =
f ( t ) dt.
(a) Find the values of g ( 2 ) and g ( −2 ) .
(b) For each of g ′( −3) and g ′′( −3) , find the value or
state that it does not exist.
(c) Find the x-coordinate of each point at which the
graph of g has a horizontal tangent line. For each
of these points, determine whether g has a relative minimum, relative maximum, or neither a minimum nor
a maximum at the point. Justify your answers.
(d) For − 4 < x < 3, find all values of x for which the graph of g has a point of inflection. Explain your
reasoning.
(a)
g ( 2) =
()
2
1
1
1
f ( t ) dt = − (1)
=−
2
2
4
1
g ( −2 ) =
−2
1
=−
f ( t ) dt = − 
1
−2
f ( t ) dt
 1 : g ( 2)
2: 
 1 : g ( −2 )
( 32 − π2 ) = π2 − 23
(b) g ′( x ) = f ( x )  g ′( −3) = f ( −3) = 2
g ′′( x ) = f ′( x )  g ′′( −3) = f ′( −3) = 1
 1 : g ′( −3)
2: 
 1 : g ′′( −3)
(c) The graph of g has a horizontal tangent line where
g ′( x ) = f ( x ) = 0. This occurs at x = −1 and x = 1.
 1 : considers g ′( x ) = 0

3 :  1 : x = −1 and x = 1
 1 : answers with justifications
g ′( x ) changes sign from positive to negative at x = −1.
Therefore, g has a relative maximum at x = −1.
g ′( x ) does not change sign at x = 1. Therefore, g has
neither a relative maximum nor a relative minimum at x = 1.
(d) The graph of g has a point of inflection at each of
x = −2, x = 0, and x = 1 because g ′′( x ) = f ′( x ) changes
sign at each of these values.
2:
{
1 : answer
1 : explanation
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AP® CALCULUS BC
2013 SCORING GUIDELINES
Question 4
The figure above shows the graph of f ′, the derivative
of a twice-differentiable function f, on the closed interval
0 ≤ x ≤ 8. The graph of f ′ has horizontal tangent lines
at x = 1, x = 3, and x = 5. The areas of the regions
between the graph of f ′ and the x-axis are labeled in the
figure. The function f is defined for all real numbers and
satisfies f ( 8 ) = 4.
(a) Find all values of x on the open interval 0 < x < 8
for which the function f has a local minimum.
Justify your answer.
(b) Determine the absolute minimum value of f on the
closed interval 0 ≤ x ≤ 8. Justify your answer.
(c) On what open intervals contained in 0 < x < 8 is the graph of f both concave down and increasing?
Explain your reasoning.
5
3
(d) The function g is defined by g ( x ) = ( f ( x ) ) . If f ( 3) = − , find the slope of the line tangent to the
2
graph of g at x = 3.
(a) x = 6 is the only critical point at which f ′ changes sign from
negative to positive. Therefore, f has a local minimum at
x = 6.
1 : answer with justification
(b) From part (a), the absolute minimum occurs either at x = 6 or
at an endpoint.
=
x 0=
and x 6
 1 : considers

3 :  1 : answer
 1 : justification
f=
( 0)
6
∫8 f ′( x ) dx
8
′( x ) dx
f (8) − ∫ f =
6
f ( 6 ) = f (8) +
=
0
∫8 f ′( x ) dx
8
f ( 8 ) − ∫ f ′( x ) dx =−
4 12 =
=
−8
0
f (8) +
4 − 7 = −3
f (8) = 4
The absolute minimum value of f on the closed interval [ 0, 8]
is −8.
(c) The graph of f is concave down and increasing on 0 < x < 1
and 3 < x < 4, because f ′ is decreasing and positive on these
intervals.
(d) g ′( x ) = 3[ f ( x )]2 · f ′( x )
( 52 ) ·4 = 75
g ′( 3=
) 3[ f ( 3)]2 · f ′( 3=
) 3 −
2
2:
{
1 : answer
1 : explanation
 2 : g ′( x )
3: 
 1 : answer
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AP® CALCULUS AB/CALCULUS BC
2014 SCORING GUIDELINES
Question 3
The function f is defined on the closed interval [ −5, 4]. The graph
of f consists of three line segments and is shown in the figure above.
Let g be the function defined by g ( x ) =
x
∫−3 f ( t ) dt.
(a) Find g ( 3) .
(b) On what open intervals contained in −5 < x < 4 is the graph
of g both increasing and concave down? Give a reason for your
answer.
(c) The function h is defined by h( x ) =
g( x)
. Find h′( 3) .
5x
(d) The function p is defined by =
p( x )
f x 2 − x . Find the slope
(
)
of the line tangent to the graph of p at the point where x = −1.
(a) g ( 3) =
3
∫−3 f ( t ) dt = 6 + 4 − 1 = 9
(b) g ′( x ) = f ( x )
2:
The graph of g is increasing and concave down on the
intervals −5 < x < −3 and 0 < x < 2 because g ′ = f is
positive and decreasing on these intervals.
(c) h′( x )
=
h′( 3) =
1 : answer
5 x g ′( x ) − g ( x ) 5 5 x g ′( x ) − 5 g ( x )
=
25 x 2
( 5 x )2
{
1 : answer
1 : reason
 2 : h′( x )
3:
 1 : answer
( 5 )( 3) g ′( 3) − 5 g ( 3)
25 ·32
15 ( −2 ) − 5 ( 9 ) −75
1
=
=
= −
225
225
3
(
)
(d) p′( x ) = f ′ x 2 − x ( 2 x − 1)
p′( −1) = f ′ ( 2 )( −3) =( −2 )( −3) = 6
 2 : p′( x )
3:
 1 : answer
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