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AP CALCULUS AB/CALCULUS BC - Tracy Unified School District Calculus AP/AP BC FR... · AP® CALCULUS AB/CALCULUS BC 2016 SCORING GUIDELINES ... x the function g ... gx f x ′′ ′

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AP® CALCULUS AB/CALCULUS BC 2016 SCORING GUIDELINES

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

Question 3

The figure above shows the graph of the piecewise-linear function f. For 4 12,x− ≤ ≤ the function g is defined by

( ) ( )2

.x

g x f t dt= ∫

(a) Does g have a relative minimum, a relative maximum, or neither at 10 ?x = Justify your answer.

(b) Does the graph of g have a point of inflection at 4 ?x = Justify your answer.

(c) Find the absolute minimum value and the absolute maximum value of g on the interval 4 12.x− ≤ ≤ Justify your answers.

(d) For 4 12,x− ≤ ≤ find all intervals for which ( ) 0.g x ≤

( ) ( ) in (a), (b), (c1 : ), r ) o (dx f xg′ =

(a) The function g has neither a relative minimum nor a

relative maximum at 10x = since ( ) ( )g x f x′ = and ( ) 0f x ≤ for 8 12.x≤ ≤

answer with justifica1 : tion

(b) The graph of g has a point of inflection at 4x = since ( ) ( )g x f x′ = is increasing for 2 4x≤ ≤ and decreasing

for 4 8.x≤ ≤

answer with justifica1 : tion

(c) ( ) ( )g x f x′ = changes sign only at 2x = − and 6.x =

x ( )g x − 4 − 4 −2 −8

6 8 12 − 4

On the interval 14 2,x− ≤ ≤ the absolute minimum value is ( )2 8g − = − and the absolute maximum value is ( )6 8.g =

1 : considers

4 : 1 : considers 4 and 12 2 : answ

2 an

ers

d 6as candida

with justificatio

tes

n

x

x x

x= = − =

− =

(d) ( ) 0g x ≤ for 24 x− ≤ ≤ and 10 12.x≤ ≤

2 : intervals

AP® CALCULUS AB/CALCULUS BC 2014 SCORING GUIDELINES

Question 3

© 2014 The College Board. Visit the College Board on the Web: www.collegeboard.org.

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 ( ) ( )3

.x

g x f t dt−

= ∫

(a) Find ( )3 .g

(b) On what open intervals contained in 5 4x− < < is the graph of g both increasing and concave down? Give a reason for your answer.

(c) The function h is defined by ( ) ( ) .5g xh x x= Find ( )3 .h′

(d) The function p is defined by ( ) ( )2 .p x f x x= − Find the slope

of the line tangent to the graph of p at the point where 1.x = −

(a) ( ) ( )3

36 4 13 9g f t dt

−= + −= =∫

1 : answer

(b) ( ) ( )fg x x′ = The graph of g is increasing and concave down on the intervals 5 3x− < < − and 0 2x< < because g f′ = is positive and decreasing on these intervals.

{ 1 : answer2 :

1 : reason

(c) ( ) ( ) ( )( )

( ) ( )2 2

5 5 5 5255

xg x g x xg x g xx

hx

x′′ ′− −

= =

( ) ( )( ) ( ) ( )

( ) ( )

25 3 3 5 3

25·315 2 5 9 75 1

225 225 3

3h g g′ −=

− − −= = = −

( ) 2 : 3 :

1 : answerh x′

(d) ( ) ( )( )2 2 1xx f x xp′ ′= − − ( ) ( )( ) ( )( )2 3 61 3 2p f ′= − = − −′ =−

( ) 2 : 3 :

1 : answerp x′

AP® CALCULUS BC 2013 SCORING GUIDELINES

Question 4

© 2013 The College Board. Visit the College Board on the Web: www.collegeboard.org.

The figure above shows the graph of ,f ′ the derivative of a twice-differentiable function f, on the closed interval 0 8.x≤ ≤ The graph of f ′ has horizontal tangent lines at 1,x = 3,x = and 5.x = 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 ( )8 4.f = (a) Find all values of x on the open interval 0 8x< <

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 8.x≤ ≤ Justify your answer.

(c) On what open intervals contained in 0 8x< < is the graph of f both concave down and increasing? Explain your reasoning.

(d) The function g is defined by ( ) ( )( )3 .g x f x= If ( ) 53 ,2f = − find the slope of the line tangent to the

graph of g at 3.x = (a) 6x = is the only critical point at which f ′ changes sign from

negative to positive. Therefore, f has a local minimum at 6.x =

1 : answer with justification

(b) From part (a), the absolute minimum occurs either at 6x = or at an endpoint.

( ) ( ) ( )

( ) ( )

0

88

04 12

0 8

88

f f

f

f x dx

f x dx

= +

′= = − = −

∫∫

( ) ( ) ( )

( ) ( )

6

88

6

6 8

8 4 7 3

f x dx

f x d

f f

f x

= ′

′= = −− −

+

=

∫∫

( )8 4f =

The absolute minimum value of f on the closed interval [ ]0, 8 is 8.−

3 : 1 : considers 0 and 6 1 : answer 1 : justification

x x= =

(c) The graph of f is concave down and increasing on 0 1x< < and 3 4,x< < because f ′ is decreasing and positive on these intervals.

2 : { 1 : answer 1 : explanation

(d) ( ) ( )[ ] ( )23 ·f x f xg x =′ ′

( ) ( )[ ] ( ) ( )22 53 3 · 33 2 753 ·4fg f= = −′ =′

3 : ( ) 2 :

1 : answerg x′

AP® CALCULUS BC 2012 SCORING GUIDELINES

Question 3

© 2012 The College Board. Visit the College Board on the Web: www.collegeboard.org.

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

be the function given by ( ) ( )1

.x

g x f t dt=

(a) Find the values of ( )2g and ( )2 .g −

(b) For each of ( )3g′ − and ( )3 ,g′′ − 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 3,x− < < find all values of x for which the graph of g has a point of inflection. Explain your reasoning.

(a) ( ) ( ) ( )( )2

12 1 1 112 2 4g f t dt = − = −=

( ) ( ) ( )

( )2 1

1 23 32 2 2

2

2

f t dt f t dtg

π π

−− =

= −

= − = −

2 : ( )( )

1 : 21 : 2

gg

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

3 3 23 3 1

g fg x f x fg x f x

g′ =

′′′ − = − =

′′ ′ ′= − = − = 2 : ( )

( )1 : 31 : 3

gg

′ − ′′ −

(c)

The graph of g has a horizontal tangent line where ( ) ( ) 0.g x f x′ = = This occurs at 1x = − and 1.x =

( )g x′ changes sign from positive to negative at 1.x = −

Therefore, g has a relative maximum at 1.x = −

( )g x′ does not change sign at 1.x = Therefore, g has neither a relative maximum nor a relative minimum at 1.x =

3 : ( ) 1 : considers 0

1 : 1 and 11 : answers with justifications

g xx x

′ = = − =

(d)

The graph of g has a point of inflection at each of 2,x = − 0,x = and 1x = because ( ) ( )g x f x′′ ′= changes

sign at each of these values.

2 : { 1 : answer1 : explanation

AP® CALCULUS BC 2011 SCORING GUIDELINES

Question 4

© 2011 The College Board. Visit the College Board on the Web: www.collegeboard.org.

The continuous function f is defined on the interval 4 3.x− ≤ ≤ The graph of f consists of two quarter circles and one line segment, as shown in the figure above.

Let ( ) ( )0

2 .x

g x x f t dt= + ∫

(a) Find ( )3 .g − Find ( )g x′ and evaluate ( )3 .g′ −

(b) Determine the x-coordinate of the point at which g has an absolute maximum on the interval 4 3.x− ≤ ≤ Justify your answer.

(c) Find all values of x on the interval 4 3x− < < 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 3.x− ≤ ≤ There is no point c, 4 3,c− < < for which ( )f c′ is equal to that average rate of change. Explain why this statement does not contradict the Mean Value Theorem.

(a) ( ) ( ) ( )3

093 63 2 4f t dtg π−

− + = − −− = ∫

( ) ( )2g x f x′ = +

( ) ( )3 2 3 2g f′ − = + − =

3 : ( )( )( )

1 : 3 1 : 1 : 3

gg xg

−⎧⎪ ′⎨⎪ ′ −⎩

(b) ( ) 0g x′ = when ( ) 2.f x = − This occurs at 5 .2x =

( ) 0g x′ > for 54 2x− < < and ( ) 0g x′ < for 5 3.2 x< <

Therefore g has an absolute maximum at 5 .2x =

3 : ( ) 1 : considers 0

1 : identifies interior candidate1 : answer with justification

g x′ =⎧⎪⎨⎪⎩

(c) ( ) ( )g x f x′′ ′= changes sign only at 0.x = Thus the graph of g has a point of inflection at 0.x =

1 : answer with reason

(d) The average rate of change of f on the interval 4 3x− ≤ ≤ is ( )3 ( 4) 2 .3 ( 4) 7

f f− −= −

− −

To apply the Mean Value Theorem, f must be differentiable at each point in the interval 4 3.x− < < However, f is not differentiable at 3x = − and 0.x =

2 : { 1 : average rate of change 1 : explanation

AP® CALCULUS BC 2010 SCORING GUIDELINES (Form B)

Question 4

© 2010 The College Board. Visit the College Board on the Web: www.collegeboard.com.

A squirrel starts at building A at time 0t = and travels along a straight wire connected to building B. For 0 18,t≤ ≤ the squirrel’s velocity is modeled by the piecewise-linear function defined by the graph above. (a) At what times in the interval 0 18,t< < if any, does the

squirrel change direction? Give a reason for your answer. (b) At what time in the interval 0 18t≤ ≤ is the squirrel

farthest from building A ? How far from building A is the squirrel at this time?

(c) Find the total distance the squirrel travels during the time interval 0 18.t≤ ≤ (d) Write expressions for the squirrel’s acceleration ( ) ,a t velocity ( ) ,v t and distance ( )x t from building A that

are valid for the time interval 7 10.t< <

(a) The squirrel changes direction whenever its velocity changes sign. This occurs at 9t = and 15.t = 2 : { 1 : -values

1 : explanationt

(b) Velocity is 0 at 0,t = 9,t = and 15.t =

t position at time t 0 0

9 9 5 20 1402+ ⋅ =

15 6 4140 10 902+− ⋅ =

18 3 290 10 1152++ ⋅ =

The squirrel is farthest from building A at time 9;t = its greatest distance from the building is 140.

2 : { 1 : identifies candidates 1 : answers

(c) The total distance traveled is ( )18

0140 50 25 215.v t dt = + + =∫

1 : answer

(d) For 7 10,t< < ( ) ( )20 10 107 10a t − −= = −

( ) ( )20 10 7 10 90v t t t= − − = − +

( ) 7 57 20 1202x += ⋅ =

( ) ( ) ( )

( )7

27

2

7 10 90

120 5 90

5 90 265

t

u t

u

x t x u du

u u

t t

=

=

= + − +

= + − +

= − + −

4 : ( )( )( )

1 : 1 :

2 :

a tv tx t

⎧⎪⎨⎪⎩

AP® CALCULUS BC 2009 SCORING GUIDELINES (Form B)

Question 5

© 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.

Let f be a twice-differentiable function defined on the interval 1.2 3.2x− < < with ( )1 2.f = The graph of ,f ′ the derivative

of f, is shown above. The graph of f ′ crosses the x-axis at 1x = − and 3x = and has a horizontal tangent at 2.x = Let g

be the function given by ( ) ( ).f xg x e=

(a) Write an equation for the line tangent to the graph of g at 1.x =

(b) For 1.2 3.2,x− < < find all values of x at which g has a local maximum. Justify your answer.

(c) The second derivative of g is ( ) ( ) ( )( ) ( )2 .f xg x e f x f x⎡ ⎤′′ ′ ′′= +⎣ ⎦ Is ( )1g′′ − positive, negative, or

zero? Justify your answer. (d) Find the average rate of change of ,g′ the derivative of g, over the interval [ ]1, 3 .

(a) ( ) ( )1 21 fg e e= =

( ) ( ) ( ) ,f xg x e f x′ ′= ( ) ( ) ( )1 21 1 4fg e f e′ ′= = −

The tangent line is given by ( )2 24 1 .y e e x= − −

3 : ( )( ) ( )

1 : 1 : 1 and 11 : tangent line equation

g xg g

′⎧⎪ ′⎨⎪⎩

(b) ( ) ( ) ( )f xg x e f x′ ′= ( ) 0f xe > for all x

So, g′ changes from positive to negative only when f ′ changes from positive to negative. This occurs at 1x = − only. Thus, g has a local maximum at 1.x = −

2 : { 1 : answer1 : justification

(c) ( ) ( ) ( )( ) ( )211 1 1fg e f f− ⎡ ⎤′′ ′ ′′− = − + −⎣ ⎦

( )1 0fe − > and ( )1 0f ′ − =

Since f ′ is decreasing on a neighborhood of 1,− ( )1 0.f ′′ − < Therefore, ( )1 0.g′′ − <

2 : { 1 : answer1 : justification

(d) ( ) ( ) ( ) ( ) ( ) ( )3 123 1 3 1 23 1 2

f fg g e f e f e′ ′ ′ ′− −= =−

2 : { 1 : difference quotient 1 : answer

AP® CALCULUS BC 2009 SCORING GUIDELINES

Question 1

© 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.

Caren rides her bicycle along a straight road from home to school, starting at home at time 0t = minutes and arriving at school at time 12t = minutes. During the time interval 0 12t≤ ≤ minutes, her velocity

( ) ,v t in miles per minute, is modeled by the piecewise-linear function whose graph is shown above.

(a) Find the acceleration of Caren’s bicycle at time 7.5t = minutes. Indicate units of measure.

(b) Using correct units, explain the meaning of ( )12

0v t dt∫ in terms of Caren’s trip. Find the value

of ( )12

0.v t dt∫

(c) Shortly after leaving home, Caren realizes she left her calculus homework at home, and she returns to get it. At what time does she turn around to go back home? Give a reason for your answer.

(d) Larry also rides his bicycle along a straight road from home to school in 12 minutes. His velocity is

modeled by the function w given by ( ) ( )sin ,15 12w t tπ π= where ( )w t is in miles per minute for

0 12t≤ ≤ minutes. Who lives closer to school: Caren or Larry? Show the work that leads to your answer.

(a) ( ) ( ) ( ) ( ) 28 77.5 7.5 0.1 miles minute8 7v va v −′= = = −− 2 : { 1 : answer

1 : units

(b) ( )12

0v t dt∫ is the total distance, in miles, that Caren rode

during the 12 minutes from 0t = to 12.t =

( ) ( ) ( ) ( )12 2 4 12

0 0 2 4

0.2 0.2 1.4 1.8 miles

v t dt v t dt v t dt v t dt= − +

= + + =∫ ∫ ∫ ∫

2 : { 1 : meaning of integral1 : value of integral

(c)

Caren turns around to go back home at time 2t = minutes. This is the time at which her velocity changes from positive to negative.

2 : { 1 : answer1 : reason

(d) ( )12

01.6;w t dt =∫ Larry lives 1.6 miles from school.

( )12

01.4;v t dt =∫ Caren lives 1.4 miles from school.

Therefore, Caren lives closer to school.

3 :

2 : Larry’s distance from school 1 : integral 1 : value1 : Caren’s distance from school

and conclusion

⎧⎪⎪⎨⎪⎪⎩

AP® CALCULUS BC 2008 SCORING GUIDELINES (Form B)

Question 5

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

Let g be a continuous function with ( )2 5.g = The graph of the piecewise-linear function ,g′ the derivative of g, is shown above for 3 7.x− ≤ ≤ (a) Find the x-coordinate of all points of inflection of the

graph of ( )y g x= for 3 7.x− < < Justify your answer.

(b) Find the absolute maximum value of g on the interval 3 7.x− ≤ ≤ Justify your answer.

(c) Find the average rate of change of ( )g x on the interval 3 7.x− ≤ ≤

(d) Find the average rate of change of ( )g x′ on the interval 3 7.x− ≤ ≤ Does the Mean Value Theorem applied on the interval 3 7x− ≤ ≤ guarantee a value of c, for 3 7,c− < < such that ( )g c′′ is equal to this average rate of change? Why or why not?

(a) g ′ changes from increasing to decreasing at 1;x = g ′ changes from decreasing to increasing at 4.x = Points of inflection for the graph of ( )y g x= occur at

1x = and 4.x =

2 : { 1 : -values1 : justification

x

(b) The only sign change of g′ from positive to negative in the interval is at 2.x =

( ) ( ) ( )( )

( ) ( ) ( )

3

2

7

2

3 153 5 5 42 22 5

1 37 5 5 4 2 2

g g x dx

g

g g x dx

−′− = + = + − + =

=

′= + = + − + =

The maximum value of g for 3 7x− ≤ ≤ is 15 .2

3 : 1 : identifies 2 as a candidate

1 : considers endpoints1 : maximum value and justification

x =⎧⎪⎨⎪⎩

(c) ( ) ( )( )

3 157 3 32 2

10 57 3g g −− − = = −− −

2 : { 1 : difference quotient 1 : answer

(d) ( ) ( )( )

( )7 3 1 4 110 27 3

g g′ ′− − − −= =− −

No, the MVT does not guarantee the existence of a value c with the stated properties because g′ is not differentiable for at least one point in 3 7.x− < <

2 : ( ) 1 : average value of 1 : answer “No” with reason

g x′⎧⎨⎩

AP® CALCULUS BC 2008 SCORING GUIDELINES

Question 4

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

A particle moves along the x-axis so that its velocity at time t, for 0 6,t≤ ≤ is given by a differentiable function v whose graph is shown above. The velocity is 0 at 0,t = 3,t = and 5,t = and the graph has horizontal tangents at 1t = and 4.t = The areas of the regions bounded by the t-axis and the graph of v on the intervals [ ]0, 3 , [ ]3, 5 , and [ ]5, 6 are 8, 3, and 2, respectively. At time 0,t = the particle is at 2.x = −

(a) For 0 6,t≤ ≤ find both the time and the position of the particle when the particle is farthest to the left. Justify your answer.

(b) For how many values of t, where 0 6,t≤ ≤ is the particle at 8 ?x = − Explain your reasoning.

(c) On the interval 2 3,t< < is the speed of the particle increasing or decreasing? Give a reason for your answer.

(d) During what time intervals, if any, is the acceleration of the particle negative? Justify your answer.

(a) Since ( ) 0v t < for 0 3t< < and 5 6,t< < and ( ) 0v t > for 3 5,t< < we consider 3t = and 6.t =

( ) ( )3

03 2 2 8 10x v t dt= − + = − − = −∫

( ) ( )6

06 2 2 8 3 2 9x v t dt= − + = − − + − = −∫

Therefore, the particle is farthest left at time 3t = when its position is ( )3 10.x = −

3 : ( )6

0

1 : identifies 3 as a candidate

1 : considers

1 : conclusion

t

v t dt

=⎧⎪⎪⎨⎪⎪⎩

(b)

The particle moves continuously and monotonically from ( )0 2x = − to ( )3 10.x = − Similarly, the particle moves

continuously and monotonically from ( )3 10x = − to ( )5 7x = − and also from ( )5 7x = − to ( )6 9.x = −

By the Intermediate Value Theorem, there are three values of t for which the particle is at ( ) 8.x t = −

3 :

1 : positions at 3, 5, and 6 1 : description of motion 1 : conclusion

t tt

= =⎧⎪ =⎪⎨⎪⎪⎩

(c) The speed is decreasing on the interval 2 3t< < since on this interval 0v < and v is increasing.

1 : answer with reason

(d) The acceleration is negative on the intervals 0 1t< < and 4 6t< < since velocity is decreasing on these intervals. 2 : { 1 : answer

1 : justification

AP® CALCULUS BC 2007 SCORING GUIDELINES (Form B)

Question 4

© 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 function defined on the closed interval 5 5x− ≤ ≤ with ( )1 3f = . The graph of ,f ′ the derivative of f, consists of two semicircles and two line segments, as shown above.

(a) For − < find all values x at which f has a relative maximum. Justify your answer.

5 x 5,<

5,<(b) For − < find all values x at which the graph of f has a point of inflection. Justify your answer.

5 x

(c) Find all intervals on which the graph of f is concave up and also has positive slope. Explain your reasoning.

(d) Find the absolute minimum value of ( )f x over the closed interval 5 x 5.− ≤ ≤ Explain your reasoning.

(a) ( ) 0f x′ = at 1, 4 3,x = −f ′ changes from positive to negative at 3− and 4.

Thus, f has a relative maximum at 3x = − and at 4.x =

2 : { 1 : -values1 : justificationx

(b) f ′ changes from increasing to decreasing, or vice versa, at and 2. Thus, the graph of f has points of

inflection when and 2. 4,x = − 1,−

4,x = − 1,−

2 : { 1 : -values1 : justificationx

(c) The graph of f is concave up with positive slope where f ′ is increasing and positive: and 1 25 4x− < < − .x< <

2 : { 1 : intervals1 : explanation

(d) Candidates for the absolute minimum are where f ′ changes from negative to positive (at 1x = ) and at the endpoints ( ). 5, 5x = −

( ) ( )5

15 3 3 22f f x dx π π

−′− = + = − + >∫ 3

( )1 3f =

( ) ( )5

13 2 15 3 3 2 2f f x dx ⋅′= + = + − >∫ 3

The absolute minimum value of f on [ ]5, 5− is ( )1 3f .=

3 : 1 : identifies 1 as a candidate 1 : considers endpoints 1 : value and explanation

x =⎧⎪⎨⎪⎩

AP® CALCULUS BC 2006 SCORING GUIDELINES (Form B)

© 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

5

Question 4

The rate, in calories per minute, at which a person using an exercise machine burns calories is modeled by the function

f. In the figure above, ( ) 3 21 3 14 2f t t t= − + + for

0 4t≤ ≤ and f is piecewise linear for 4 24.t≤ ≤

(a) Find ( )22 .f ′ Indicate units of measure.

(b) For the time interval 0 24,t≤ ≤ at what time t is f increasing at its greatest rate? Show the reasoning that supports your answer.

(c) Find the total number of calories burned over the time interval 6 18t≤ ≤ minutes.

(d) The setting on the machine is now changed so that the person burns ( )f t c+ calories per minute. For this setting, find c so that an average of 15 calories per minute is burned during the time interval 6 18.t≤ ≤

(a) ( ) 15 322 320 24f −′ = = −−

calories/min/min

1 : ( )22f ′ and units

(b) f is increasing on [ ]0, 4 and on [ ]12, 16 .

On ( )12, 16 , ( ) 15 9 316 12 2f t −′ = =

− since f has

constant slope on this interval.

On ( )0, 4 , ( ) 23 34f t t t′ = − + and

( ) 3 3 02f t t′′ = − + = when 2.t = This is where f ′

has a maximum on [ ]0, 4 since 0f ′′ > on ( )0, 2 and 0f ′′ < on ( )2, 4 .

On [ ]0, 24 , f is increasing at its greatest rate when

2t = because ( ) 32 3 .2f ′ = >

4 :

( )( )

( ) ( )

1 : on 0, 41 : shows has a max at 2 on 0, 41 : shows for 12 16, 2

1 : answer

ff t

t f t f

′⎧⎪ ′ =⎪⎨ ′ ′< < <⎪⎪⎩

(c) ( ) ( ) ( )( ) ( )18

616 9 4 9 15 2 152

132 calories

f t dt = + + +

=∫

2 : { 1 : method1 : answer

(d) We want ( )( )18

61 15.12 f t c dt+ =∫

This means 132 12 15(12).c+ = So, 4.c =

OR

Currently, the average is 132 1112 = calories/min.

Adding c to ( )f t will shift the average by c. So 4c = to get an average of 15 calories/min.

2 : { 1 : setup1 : value of c

AP® CALCULUS BC 2005 SCORING GUIDELINES (Form B)

Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

5

Question 4

The graph of the function f above consists of three line segments.

(a) Let g be the function given by ( ) ( )4

.x

g x f t dt−

= ∫

For each of ( )1 ,g − ( )1 ,g′ − and ( )1 ,g′′ − find the value or state that it does not exist.

(b) For the function g defined in part (a), find the x-coordinate of each point of inflection of the graph of g on the open interval 4 3.x− < < Explain your reasoning.

(c) Let h be the function given by ( ) ( )3

.x

h x f t dt= ∫ Find all values of x in the closed interval

4 3x− ≤ ≤ for which ( ) 0.h x =

(d) For the function h defined in part (c), find all intervals on which h is decreasing. Explain your reasoning.

(a) ( ) ( ) ( )( )1

41 151 3 52 2g f t dt

−− = = − = −∫

( ) ( )1 1 2g f′ − = − = − ( )1g′′ − does not exist because f is not differentiable

at 1.x = −

3 : ( )( )( )

1 : 11 : 11 : 1

ggg

−⎧⎪ ′ −⎨⎪ ′′ −⎩

(b) 1x = g f′ = changes from increasing to decreasing at 1.x =

2 : 1 : 1 (only)

1 : reasonx =⎧

⎨⎩

(c) 1, 1, 3x = − 2 : correct values 1− each missing or extra value

(d) h is decreasing on [ ]0, 2 0h f′ = − < when 0f >

2 : 1 : interval1 : reason

⎧⎨⎩

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6

Question 5

A car is traveling on a straight road. For 0 24t≤ ≤ seconds, the car’s velocity ( ) ,v t in meters per second, is modeled by the piecewise-linear function defined by the graph above.

(a) Find ( )24

0.v t dt∫ Using correct units, explain the meaning of ( )

24

0.v t dt∫

(b) For each of ( )4v′ and ( )20 ,v′ find the value or explain why it does not exist. Indicate units of measure.

(c) Let ( )a t be the car’s acceleration at time t, in meters per second per second. For 0 24,t< < write a piecewise-defined function for ( ).a t

(d) Find the average rate of change of v over the interval 8 20.t≤ ≤ Does the Mean Value Theorem guarantee a value of c, for 8 20,c< < such that ( )v c′ is equal to this average rate of change? Why or why not?

(a) ( ) ( )( ) ( )( ) ( )( )24

01 14 20 12 20 8 20 3602 2v t dt = + + =∫

The car travels 360 meters in these 24 seconds.

2 : { 1 : value1 : meaning with units

(b) ( )4v′ does not exist because ( ) ( ) ( ) ( )

4 4

4 4lim 5 0 lim .4 4t t

v t v v t vt t− +→ →

− −⎛ ⎞ ⎛ ⎞= ≠ =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

( ) 220 0 520 m sec16 24 2v −′ = = −−

3 : ( )( )

1 : 4 does not exist, with explanation 1 : 20 1 : units

vv′⎧

⎪ ′⎨⎪⎩

(c)

( )

5 if 0 4 0 if 4 16

5 if 16 242

tta tt

< <⎧⎪ < <= ⎨⎪− < <⎩

( )a t does not exist at 4t = and 16.t =

2 : 5 1 : finds the values 5, 0, 2

1 : identifies constants with correct intervals

⎧ −⎪⎨⎪⎩

(d) The average rate of change of v on [ ]8, 20 is ( ) ( ) 220 8 5 m sec .20 8 6

v v−= −

No, the Mean Value Theorem does not apply to v on [ ]8, 20 because v is not differentiable at 16.t =

2 : [ ]1 : average rate of change of on 8, 20 1 : answer with explanation

v⎧⎨⎩

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5

Question 4

The figure above shows the graph of ,f ′ the derivative of the function f, on the closed interval 1 5.x− ≤ ≤ The graph of f ′ has horizontal tangent lines at 1x = and 3.x = The function f is twice differentiable with

( )2 6.f = (a) Find the x-coordinate of each of the points of inflection of the graph

of f. Give a reason for your answer. (b) At what value of x does f attain its absolute minimum value on the

closed interval 1 5 ?x− ≤ ≤ At what value of x does f attain its absolute maximum value on the closed interval 1 5 ?x− ≤ ≤ Show the analysis that leads to your answers.

(c) Let g be the function defined by ( ) ( ).g x x f x= Find an equation for the line tangent to the graph of g at 2.x =

(a) 1x = and 3x = because the graph of f ′ changes from

increasing to decreasing at 1,x = and changes from decreasing to increasing at 3.x =

2 : 1 : 1, 3

1 : reasonx x= =

(b) The function f decreases from 1x = − to 4,x = then increases from 4x = to 5.x = Therefore, the absolute minimum value for f is at 4.x =

The absolute maximum value must occur at 1x = − or at 5.x =

( ) ( ) ( )5

15 1 0f f f t dt

−′− − = <∫

Since ( ) ( )5 1 ,f f< − the absolute maximum value occurs at 1.x = −

4 :

1 : indicates decreases then increases 1 : eliminates 5 for maximum 1 : absolute minimum at 4 1 : absolute maximum at 1

fx

xx

= = = −

(c) ( ) ( ) ( )g x f x x f x′ ′= + ( ) ( ) ( ) ( )2 2 2 2 6 2 1 4g f f′ ′= + = + − = ( ) ( )2 2 2 12g f= = Tangent line is ( )4 2 12y x= − +

3 : ( ) 2 :

1 : tangent lineg x′

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6

Question 5

Let f be a function defined on the closed interval [0,7]. The graph of

f, consisting of four line segments, is shown above. Let g be the

function given by 2

( ) ( ) .x

g x f t dt=

(a) Find ( )3 ,g ( )3 ,g and ( )3 .g

(b) Find the average rate of change of g on the interval 0 3.x

(c) For how many values c, where 0 3,c< < is ( )g c equal to the

average rate found in part (b)? Explain your reasoning.

(d) Find the x-coordinate of each point of inflection of the graph of

g on the interval 0 7.x< < Justify your answer.

(a) ( )3

2

1(3) ( ) 4 2 3

2g f t dt= = + =

(3) (3) 2g f= =

0 4

(3) (3) 24 2

g f= = =

3 :

1 : (3)

1 : (3)

1 : (3)

g

g

g

(b) (3) (0)

3

g g =

3

0

1( )

3f t dt

= ( )1 1 1 7(2)(4) (4 2)

3 2 2 3+ + =

2 :

3

01 : (3) (0) ( )

1 : answer

g g f t dt=

(c) There are two values of c.

We need 7

( ) ( )3

g c f c= =

The graph of f intersects the line 7

3y = at two

places between 0 and 3.

2 : 1 : answer of 2

1 : reason

Note: 1/2 if answer is 1 by MVT

(d) x = 2 and x = 5

because g f= changes from increasing to

decreasing at x = 2, and from decreasing to

increasing at x = 5.

2 :

1 : 2 and 5 only

1 : justification

(ignore discussion at 4)

x x

x

= =

=

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2003 SCORING GUIDELINES

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5

Question 4

Let f be a function defined on the closed interval 3 4x with

( )0 3.f = The graph of ,f the derivative of f, consists of one line

segment and a semicircle, as shown above.

(a) On what intervals, if any, is f increasing? Justify your answer.

(b) Find the x-coordinate of each point of inflection of the graph of f

on the open interval 3 4.x< < Justify your answer.

(c) Find an equation for the line tangent to the graph of f at the

point ( )0,3 .

(d) Find ( )3f and ( )4 .f Show the work that leads to your answers.

(a) The function f is increasing on [ 3, 2] since

0f > for 3 2x < . 2 :

1 : interval

1 : reason

(b) 0x = and 2x =

f changes from decreasing to increasing at

0x = and from increasing to decreasing at

2x =

2 : 1 : 0 and 2 only

1 : justification

x x= =

(c) (0) 2f =

Tangent line is 2 3.y x= +

1 : equation

(d) (0) ( 3)f f 0

3( )f t dt=

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

2 2 2= =

3 9

( 3) (0)2 2

f f= + =

(4) (0)f f 4

0( )f t dt=

( )218 (2) 8 2

2= = +

(4) (0) 8 2 5 2f f= + = +

4 :

( )

( )2

1 1 : 2

2

(difference of areas

of triangles)

1 : answer for ( 3) using FTC

1 1 : 8 (2)

2

(area of rectangle

area of semicircle)

1 : answer for (4) using FTC

f

f

±

±

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5

Question 4

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5

Question 4 �

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Question 3

A car is traveling on a straight road with velocity

55 ft/sec at time t = 0. For 0 18t� � seconds, the

car�s acceleration ( )a t , in ft/sec2, is the piecewise

linear function defined by the graph above.

(a) Is the velocity of the car increasing at t = 2

seconds? Why or why not?

(b) At what time in the interval 0 18t� � , other than t = 0, is the velocity of the car

55 ft/sec? Why?

(c) On the time interval 0 18t� � , what is the car�s absolute maximum velocity, in ft/sec,

and at what time does it occur? Justify your answer.

(d) At what times in the interval 0 18t� � , if any, is the car�s velocity equal to zero? Justify

your answer.

(a) Since (2) (2)v a� � and (2) 15 0a � � , the velocity is

increasing at t = 2.

1 : answer and reason

(b) At time t = 12 because 12

0(12) (0) ( ) 0v v a t dt� � �� .

2 : 1 : 12

1 : reason

t ��������

(c) The absolute maximum velocity is 115 ft/sec at

t = 6.

The absolute maximum must occur at t = 6 or

at an endpoint.

6

0(6) 55 ( )

155 2(15) (4)(15) 115 (0)2

v a t dt

v

� �

� � � � �

18

6( ) 0a t dt �� so (18) (6)v v�

4 :

1 : 6

1 : absolute maximum velocity

1 : identifies 6 and

18 as candidates or

indicates that increases,

decreases, then increases

1 : eliminates 18

t

t

t

v

t

���������� ����� �����������

����������

(d) The car�s velocity is never equal to 0. The absolute

minimum occurs at t = 16 where 16

6(16) 115 ( ) 115 105 10 0v a t dt� � � � � �� .

2 : 1 : answer

1 : reason

�������

Copyright © 2000 by College Entrance Examination Board and Educational Testing Service. All rights reserved.AP is a registered trademark of the College Entrance Examination Board.

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t1 2 3 4 5 6 7 8 9 10

123456789

1011121314

0

Time(seconds)

Vel

ocity

of

Run

ner

A(m

eter

s pe

r se

cond

) (3, 10) (10, 10)

AB{5 / BC{5 1999

5. The graph of the function f , consisting of three line segments, is

given above. Let g(x) =

Zx

1

f(t) dt.

(a) Compute g(4) and g(�2).

(b) Find the instantaneous rate of change of g, with respect to x, atx = 1.

(c) Find the absolute minimum value of g on the closed interval[�2; 4]. Justify your answer.

(d) The second derivative of g is not de�ned at x = 1 and x = 2.How many of these values are x{coordinates of points ofin ection of the graph of g? Justify your answer.

O

(1, 4)

(2, 1)

(4, –1)

1–1 2–2 3 4

1

–1

2

–2

3

4

(a) g(4) =

Z4

1

f(t) dt =3

2+ 1 +

1

2�

1

2=

5

2

g(�2) =

Z�2

1

f(t) dt = �1

2(12) = �6

2

(1: g(4)

1: g(�2)

(b) g0(1) = f(1) = 4 1: answer

(c) g is increasing on [�2; 3] and decreasing on [3; 4].

Therefore, g has absolute minimum at anendpoint of [�2; 4].

Since g(�2) = �6 and g(4) =5

2,

the absolute minimum value is �6.

3

8>><>>:

1: interior analysis

1: endpoint analysis

1: answer

(d) One; x = 1

On (�2; 1), g00(x) = f 0(x) > 0

On (1; 2), g00(x) = f 0(x) < 0

On (2; 4), g00(x) = f 0(x) < 0

Therefore (1; g(1)) is a point of in ection and(2; g(2)) is not.

3

8>><>>:

1: choice of x = 1 only

1: show (1; g(1)) is a point of in ection

1: show (2; g(2)) is not a point of in ection