2017 AP® CALCULUS AB FREE-RESPONSE QUESTIONS

6. Let f be the function defined by sin xf x cos 2 x � e ( ) ( ) .

Let g be a differentiable function. The table above gives values of g and its derivative g � at selected values of x.

Let h be the function whose graph, consisting of five line segments, is shown in the figure above.

(a) Find the slope of the line tangent to the graph of f at x p .

(b) Let k be the function defined by k x h f x( ) ( ( )) .. Find k �(p )

(c) Let m be the function defined by m x( ) ( g −2x ¹ h x) ( ). Find m�(2).

(d) Is there a number c in the closed interval > 5, 3 − − @ � 4g c( ) − such that ? Justify your answer.

STOP

END OF EXAM

© 2017 The College Board. Visit the College Board on the Web: www.collegeboard.org. 

-7-

AP® CALCULUS AB 2017 SCORING GUIDELINES

© 2017 The College Board. Visit the College Board on the Web: www.collegeboard.org.

Question 6

(a) ( ) ( ) sin2sin 2 cos xf x x x e′ = − +

( ) ( ) sin2sin 2 cos 1f e′ = − + = −

( )2 : f ′

(b) ( ) ( )( ) ( )k x h f x f x′ ′= ′ ( ) ( )( ) ( ) ( ) ( )

( )( )

2 11 113 3

k h f f h′ ′ ′ ′= = −

= − − =

( )( )

1 : 2 :

1 : k xk′′

(c) ( ) ( ) ( ) ( ) ( )2 2 2m x g x h x g x h x′ ′ − + − ′= −

( ) ( ) ( ) ( ) ( )

( )( ) ( )2 2 4 2 4 2

2 12 1 5 33 3

m g h g h′ ′ ′= − − + −

= − − − + − = −

( )( )

2 : 2

3 : 1 :

m xm′′

(d) g is differentiable. g is continuous on the interval [ 5, 3].− − ( ) ( )

( )3 5 2 10 423 5

g g− − − −= = −− − −

Therefore, by the Mean Value Theorem, there is at least one value c,

5 3,c− < < − such that ( ) 4.g c′ = −

( ) ( )( ) 1 :

2 : 1 : justification, using Mean Value The

3 5

or

5

em

3gg −− −

− − −

AP® CALCULUS AB 2016 SCORING GUIDELINES

© 2016 The College Board. Visit the College Board on the Web: www.collegeboard.org.

Question 6

x ( )f x ( )f x′ ( )g x ( )g x′

1 – 6 3 2 8

2 2 –2 –3 0

3 8 7 6 2

6 4 5 3 –1

The functions f and g have continuous second derivatives. The table above gives values of the functions and their derivatives at selected values of x.

(a) Let ( ) ( )( ).k x f g x= Write an equation for the line tangent to the graph of k at 3.x =

(b) Let ( ) ( )( ) .g xh x f x= Find ( )1 .h′

(c) Evaluate ( )3

12 .f x dx′′∫

(a) ( ) ( )( ) ( )

( ) ( )( ) ( ) ( )3 3

3

3 6 2 5 2 1

6 4

0

3

k f gk f g f

g f= = =

′ ′ ′ ′= = = =

An equation for the tangent line is ( )10 3 4.y x= − +

{ 2 : slope at 33 :

1 : equation for tangent line

x =

(b) ( ) ( ) ( ) ( ) ( )( )( )

( )( )

2

2

1 1 1 11

1

6 8 2 3 54 336 26

g ff

h f g′ ′−=

−

′

− −=−

= = −

( ){ 2 : expression for 3 :

1 : a er

1

nsw

h′

(c) ( ) ( ) ( ) ( )

( )

3 3

11

1 12 2 6 22 21 75 22 2

f x dx f x f f=′′ ′ ′ ′= −

= − − =

∫

{ 2 : antiderivative3 :

1 : answer

AP® CALCULUS AB 2008 SCORING GUIDELINES (Form B)

Question 4

© 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.

The functions f and g are given by ( )3 20

4x

f x t dt= +³ and ( ) ( )sin .g x f x=

(a) Find ( )f x′ and ( ).g x′

(b) Write an equation for the line tangent to the graph of ( )y g x= at .x π=

(c) Write, but do not evaluate, an integral expression that represents the maximum value of g on the interval 0 .x π≤ ≤ Justify your answer.

(a) ( ) ( )23 4 3f x x′ = +

( ) ( )

( )2

sin cos

3 4 3sin cos

g x f x x

x x

′ ′= ⋅

= + ⋅

4 : ( )( )

2 : 2 :

f xg x

′­® ′¯

(b) ( ) 0,g π = ( ) 6g π′ = − Tangent line: ( )6y x π= − −

2 : ( ) ( ) 1 : or 1 : tangent line equation

g gπ π′­®¯

(c) For 0 ,x π< < ( ) 0g x′ = only at .2x π=

( ) ( )0 0g g π= =

( ) 3 20

4 02g t dtπ = + >³

The maximum value of g on [ ]0, π is 3 20

4 .t dt+³

3 :

( )

( )

1 : sets 0

1 : justifies maximum at 2

1 : integral expression for 2

g x

g

π

π

′ =­°°®°°̄

126

AP® CALCULUS AB 2007 SCORING GUIDELINES

Question 3

© 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).

x ( )f x ( )f x′ ( )g x ( )g x′

1 6 4 2 5

2 9 2 3 1

3 10 – 4 4 2

4 –1 3 6 7

The functions f and g are differentiable for all real numbers, and g is strictly increasing. The table above gives values of the functions and their first derivatives at selected values of x. The function h is given by ( ) ( )( ) 6.h x f g x= −

(a) Explain why there must be a value r for 1 3r< < such that ( ) 5.h r = −

(b) Explain why there must be a value c for 1 3c< < such that ( ) 5.h c′ = −

(c) Let w be the function given by ( ) ( )( )

1.

g xw x f t dt= ³ Find the value of ( )3 .w′

(d) If 1g− is the inverse function of g, write an equation for the line tangent to the graph of ( )1y g x−= at 2.x =

(a) ( ) ( )( ) ( )1 1 6 2 6 9 6 3h f g f= − = − = − = ( ) ( )( ) ( )3 3 6 4 6 1 6 7h f g f= − = − = − − = −

Since ( ) ( )3 5 1h h< − < and h is continuous, by the Intermediate Value Theorem, there exists a value r, 1 3,r< < such that ( ) 5.h r = −

2 : ( ) ( ) 1 : 1 and 31 : conclusion, using IVT

h h­®¯

(b) ( ) ( )3 1 7 3 53 1 3 1h h− − −= = −− −

Since h is continuous and differentiable, by the Mean Value Theorem, there exists a value c, 1 3,c< < such that ( ) 5.h c′ = −

2 : ( ) ( )3 1 1 : 3 1

1 : conclusion, using MVT

h h−­° −®°̄

(c) ( ) ( )( ) ( ) ( )3 3 3 4 2 2w f g g f′ ′= ⋅ = ⋅ = − 2 : { 1 : apply chain rule 1 : answer

(d) ( )1 2,g = so ( )1 2 1.g− =

( ) ( )( )( ) ( )

11

1 1 12 512g

gg g−

−′ = = =′′

An equation of the tangent line is ( )11 2 .5y x− = −

3 :

( )

( ) ( )

1

1

1 : 2

1 : 2

1 : tangent line equation

g

g

−

−

­°° ′®°°̄

108

AP® CALCULUS AB 2007 SCORING GUIDELINES (Form B)

Question 6

© 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).

Let f be a twice-differentiable function such that � �2f 5 and � �5 2f . Let g be the function given by

� � � �� �.g x f f x

(a) Explain why there must be a value c for 2 5c� � such that � � 1.f c �c

(b) Show that Use this result to explain why there must be a value k for 2 such that � � � �2g g c c 5 . 5k� �� � 0.g k cc

(c) Show that if � � 0f x cc for all x, then the graph of g does not have a point of inflection.

(d) Let � � � � .h x f x x � Explain why there must be a value r for 2 5r� � such that � � 0.h r

(a) The Mean Value Theorem guarantees that there is a value with so that ,c 2 c� � 5,

� � � � � �5 2 2 5 1.5 2 5 2

f ff c � �c � �

�

2 : � � � �5 2

1 : 5 2

1 : conclusion, using MVT

f f�­°�®

°̄

(b) � � � �� � � �g x f f x f xc c c �

� � � �� � � � � � � �2 2 2 5g f f f f fc c c c c � � 2

� � � �� � � � � � � �5 5 5 2g f f f f fc c c c c � � 5

Thus, � � � �2 5g gc c .

Since f is twice-differentiable, gc is differentiable

everywhere, so the Mean Value Theorem applied to gc on

guarantees there is a value k, with >2, 5@ ,2 5k� � such

that � � � � � �5 20.

5 2g gg kc c�cc

�

3 :

� �� � � � � � � �

1 :

1 : 2 5 2 5

1 : uses MVT with

g xg f f g

g

c­° c c c c � ®° c¯

(c) � � � �� � � � � � � �� � � �g x f f x f x f x f f x f xcc cc c c c cc � � � �

If � � 0f xcc for all x, then

� � � � � � � �� �0 0g x f x f x f f xcc c c c � � � � 0 for all x. Thus, there is no x-value at which � �g xcc changes sign, so

the graph of g has no inflection points.

2 : � �

1 : considers

1 : 0 for all

gg x x

cc­® cc ¯

OR OR

2 : ^ 1 : is linear

1 : is linear

fg

If � � 0f xcc for all x, then f is linear, so g f f D is

linear and the graph of g has no inflection points.

(d) Let � � � � .h x f x x �

� � � �� � � �2 2 2 5 2 3

5 5 5 2 5

h fh f

� � � � 3�

Since � � � �2 0 5h h! ! ,

5, the Intermediate Value Theorem

guarantees that there is a value r, with 2 r� � such that � � 0.h r

2 : � � � � 1 : 2 and 5

1 : conclusion, using IVT

h h­®¯

121

AP® CALCULUS AB 2006 SCORING GUIDELINES

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7

Question 6 The twice-differentiable function f is defined for all real numbers and satisfies the following conditions:

( )0 2,f = ( )0 4,f = −′ and ( )0 3.f =′′

(a) The function g is given by ( ) ( )axg x e f x= + for all real numbers, where a is a constant. Find ( )0g ′ and ( )0g ′′ in terms of a. Show the work that leads to your answers.

(b) The function h is given by ( ) ( ) ( )cosh x kx f x= for all real numbers, where k is a constant. Find ( )h x′ and write an equation for the line tangent to the graph of h at 0.x =

(a) ( ) ( )axg x ae f x′ ′= + ( )0 4g a′ = −

( ) ( )2 axg x a e f x′′ ′′= +

( ) 20 3g a′′ = +

4 :

( )( )( )( )

1 : 1 : 0

1 : 1 : 0

g xgg xg

′­° ′°® ′′°° ′′¯

(b) ( ) ( ) ( ) ( ) ( )cos sinh x f x kx k kx f x′ ′= − ( ) ( ) ( ) ( ) ( ) ( )0 0 cos 0 sin 0 0 0 4h f k f f′ ′ ′= − = = − ( ) ( ) ( )0 cos 0 0 2h f= =

The equation of the tangent line is 4 2.y x= − +

5 :

( )( )( )

2 : 1 : 0

3 : 1 : 01 : equation of tangent line

h xhh

′­° ′° ­

°®®°°°̄ ¯

120

AP® CALCULUS AB 2005 SCORING GUIDELINES

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5

Question 4

x 0 0 1x< < 1 1 2x< < 2 2 3x< < 3 3 4x< < ( )f x –1 Negative 0 Positive 2 Positive 0 Negative ( )f x′ 4 Positive 0 Positive DNE Negative –3 Negative ( )f x′′ –2 Negative 0 Positive DNE Negative 0 Positive

Let f be a function that is continuous on the interval [ )0, 4 . The function f is twice differentiable except at 2.x = The function f and its derivatives have the properties indicated in the table above, where DNE indicates that the derivatives of f do not exist at 2.x = (a) For 0 4,x< < find all values of x at which f has a relative extremum. Determine whether f has a relative maximum

or a relative minimum at each of these values. Justify your answer. (b) On the axes provided, sketch the graph of a function that has all the characteristics of f.

(Note: Use the axes provided in the pink test booklet.)

(c) Let g be the function defined by ( ) ( )1

xg x f t dt= ³ on the open interval ( )0, 4 . For

0 4,x< < find all values of x at which g has a relative extremum. Determine whether g has a relative maximum or a relative minimum at each of these values. Justify your answer.

(d) For the function g defined in part (c), find all values of x, for 0 4,x< < at which the graph of g has a point of inflection. Justify your answer.

(a) f has a relative maximum at 2x = because f ′ changes from positive to negative at 2.x =

2 : { 1 : relative extremum at 21 : relative maximum with justification

x =

(b)

2 : ( )

1 : points at 0, 1, 2, 3 and behavior at 2, 21 : appropriate increasing/decreasing

and concavity behavior

x =­°°®°°̄

(c) ( ) ( ) 0g x f x= =′ at 1, 3.x = g′ changes from negative to positive at 1x = so g has a relative

minimum at 1.x = g′ changes from positive to negative at 3x = so g has a relative maximum at 3.x =

3 : ( ) ( ) 1 :

1 : critical points1 : answer with justification

g x f x′ =­°®°̄

(d) The graph of g has a point of inflection at 2x = because g f′′ ′= changes sign at 2.x =

2 : { 1 : 21 : answer with justification

x =

107

AP® CALCULUS AB 2002 SCORING GUIDELINES

Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

7

Question 6 !

!" !"#! " !"$! $"#! !" !"# " !"# " !"$ "

! "! " " !! " $! " %! " !! " %! " $! " !! "

! "! "! " !! " $! " %! " $ " % " # " # "

!"#$ #$ %"$&$'()*#+,)$#-&#$+.$/+''"0")#+&%1"$',0$&11$0"&1$)(2%"0.3$4-"$#&%1"$&%,5"$6+5".$#-"$5&1(".$,'$ #$ &)/$+#.$

/"0+5&#+5"$ ! ! $',0$."1"*#"/$7,+)#.$ !$ +)$#-"$*1,."/$+)#"05&1$ !"# !"#"!! " " $4-"$."*,)/$/"0+5&#+5"$,'$ #$ -&.$#-"$

70,7"0#8$#-&#$ ! " #! "!! " $',0$ !"# !"#"!! " " $

9&:$ ;5&1(&#"$ ! "!"#

$

! " # $ %! " #"! "! $<-,=$#-"$=,0>$#-&#$1"&/.$#,$8,(0$&).="03$

9%:$ ?0+#"$&)$"@(&#+,)$,'$#-"$1+)"$#&)6")#$#,$#-"$60&7-$,'$ #$ &#$#-"$7,+)#$=-"0"$ !"! ! $A."$#-+.$1+)"$#,$

&770,B+2&#"$#-"$5&1("$,'$ !$%&"%! $C.$#-+.$&770,B+2&#+,)$60"&#"0$#-&)$,0$1"..$#-&)$#-"$&*#(&1$5&1("$,'$ !$%&"'! $

D+5"$&$0"&.,)$',0$8,(0$&).="03$

9*:$ E+)/$&$7,.+#+5"$0"&1$)(2%"0$ $$ -&5+)6$#-"$70,7"0#8$#-&#$#-"0"$2(.#$"B+.#$&$5&1("$ %$ =+#-$ $ $"#"# # $&)/$$

! " #! " #!! " $D+5"$&$0"&.,)$',0$8,(0$&).="03$

9/:$ !"#$ &$ %"$#-"$'()*#+,)$6+5")$%8$

!

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! ! !

" !

! ! !

! " " #$$$% &$ ' " ($$)

$

$ 4-"$60&7-$,'$ &$ 7&..".$#-0,(6-$"&*-$,'$#-"$7,+)#.$! "! " #! " ! $6+5")$+)$#-"$#&%1"$&%,5"3$C.$+#$7,..+%1"$#-&#$#$ &)/$

&$ &0"$#-"$.&2"$'()*#+,)F$D+5"$&$0"&.,)$',0$8,(0$&).="03!!

9&:$ ! "!"# !"# !"#

$ $ $

! " # $ ! " # $! " #" ! " #" #"! !" # "! ! ! $

$ $ ! " ! "!"#

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! " # ! $ " %# & $'( )&!! " !! " ! # # # " ! $

G$$ ( )*+,-./0-1*,-1/

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!9%:$ ! "! " #! "! " " $

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60&7-$,'$#$+.$*,)*&5"$(7$,)$#-"$+)#"05&1$

! !"#!! ! 3$

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! " #+,-./$(0# #,&#$%&'(&$#)*&(#%$# !12

! " #%&03(4#3*$5#4(%0,&

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$ $"%"! ! $.(*-$#-&#$

$!"#$% !"% & "

! % '"#$ " "#$

! !! " #

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! " #4(6(4(&+(#$,#789#6,4# #:,4#;*66(4(&$*%<*)*$=

######,6# >#

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#

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!"# $ % !"# & ' '! !

! " "! !

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! " # " # $

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!"# $ % !"# & ' '! !

! " "! !

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$ 4-(.$! ! $+.$),#$*,)#+)(,(.$&#$!$L$MN$%(#$ " ! $+.$

*,)#+)(,(.$&#$!$L$MN$.,$ ! "! 3$

$ $ $ $ OP$

$ ! " #! "!! " $',0$&11$ !! ! N$%(#$+#$=&.$.-,=)$+)$7&0#$

9*:$#-&#$ ! " #! "!! " $',0$.,2"$ !! ! N$.,$ ! "! 3$

$$

G$

! " #$%&'()&#*%+,#'-./#)(0()(%1(#.+#

###### #+)# #

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111

1991 AB5

Let f be a function that is even and continuous on the closed interval [ . The function f and its derivatives have the properties indicated in the table below.

−3,3]

x 0 0 < x <1 1 1 < x < 2 2 2 < x < 3f (x) 1 Positive 0 Negative −1 Negative′ f (x) Undefined Negative 0 Negative Undefined Positive′ ′ f (x) Undefined Positive 0 Negative Undefined Negative

(a) Find the x-coordinate of each point at which f attains an absolute maximum value

or an absolute minimum value. For each x-coordinate you give, state whether f attains an absolute maximum or an absolute minimum.

(b) Find the x-coordinate of each point of inflection on the graph of f . Justify your

answer. (c) In the xy-plane provided below, sketch the graph of a function with all the given

characteristics of f .

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

y

x

Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 109

1991 AB5 Solution

(a) Absolute maximum at 0x =Absolute minimum at 2x = ±

(b) Points of inflection at because the sign of 1x = ± ( )f x′′ changes at 1x = and f is even

−3 −2 −1 1 2 3

−1

1

y

x

(c)

Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 110