AP CALCULUS AB/CALCULUS BC 2016 SCORING GUIDELINES

Size: px

Start display at page:

Download "AP CALCULUS AB/CALCULUS BC 2016 SCORING GUIDELINES"

Allen Dawson
6 years ago
Views:

1 The figure above shows the graph of the piecewise-linear function f. For 4, the function g is defined by g( ) = f ( t) (a) Does g have a relative minimum, a relative maimum, or neither at =? Justify your answer. (b) Does the graph of g have a point of inflection at = 4? Justify your answer. (c) Find the absolute minimum value and the absolute maimum value of g on the interval 4. Justify your answers. AP CALCULUS AB/CALCULUS BC 6 SCORING GUIDELINES Question (d) For 4, find all intervals for which g( ). : g ( ) = f ( ) in (a), (b), (c), o r (d) (a) The function g has neither a relative minimum nor a relative maimum at = since g ( ) = f ( ) and f ( ) for 8. (b) The graph of g has a point of inflection at = 4 since g ( ) = f ( ) is increasing for 4 and decreasing for 4 8. (c) g ( ) = f ( ) changes sign only at = and = 6. g( ) : answer with justification : answer with justification : considers = and = 6 as candidates 4: : considers = 4 and = : answers with justification On the interval 4, the absolute minimum value is g( ) = 8 and the absolute maimum value is g ( 6) = 8. (d) g( ) for 4 and. : intervals 6 The College Board. Visit the College Board on the Web:

2 The function f is defined on the closed interval [ 5, 4 ]. AP CALCULUS AB/CALCULUS BC 4 SCORING GUIDELINES Question The graph of f consists of three line segments and is shown in the figure above. Let g be the function defined by g( ) = f ( t) (a) Find g (. ) (b) On what open intervals contained in 5 < < 4 is the graph of g both increasing and concave down? Give a reason for your answer. g( ) (c) The function h is defined by h ( ) =. Find h (. ) 5 (d) The function p is defined by p ( ) f( ) =. Find the slope of the line tangent to the graph of p at the point where =. (a) g( ) = f ( t) dt = = 9 : answer (b) g ( ) = f ( ) The graph of g is increasing and concave down on the intervals 5 < < and < < because g = f is positive and decreasing on these intervals. : { : answer : reason 5g ( ) g( ) 5 5g ( ) 5g( ) (c) h ( ) = = ( 5) 5 : h ( ) : : answer ( 5)( ) g ( ) 5g( ) h ( ) = 5 5( ) 5( 9) 75 = = = 5 5 (d) p ( ) = f ( )( ) p ( ) = f ( )( ) = ( )( ) = 6 : p ( ) : : answer 4 The College Board. Visit the College Board on the Web:

3 SCORING GUIDELINES Question Let f be the continuous function defined on [ 4, ] whose graph, consisting of three line segments and a semicircle centered at the origin, is given above. Let g be the function given by g( ) = f( t) (a) Find the values of g ( ) and g(. ) (b) For each of g ( ) and g (, ) find the value or state that it does not eist. (c) Find the -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 maimum, or neither a minimum nor a maimum at the point. Justify your answers. (d) For 4 < <, find all values of for which the graph of g has a point of inflection. Eplain your reasoning. (a) g( ) = f( t) dt = ()( ) = 4 : g( ) : : g( ) g( ) = f () t dt = f() t dt π π = ( ) = (b) g ( ) = f( ) g ( ) = f( ) = g ( ) = f ( ) g ( ) = f ( ) = : : g ( ) : g ( ) (c) The graph of g has a horizontal tangent line where g ( ) = f( ) =. This occurs at = and =. g ( ) changes sign from positive to negative at =. Therefore, g has a relative maimum at =. : considers g ( ) = : : = and = : answers with justifications g ( ) does not change sign at =. Therefore, g has neither a relative maimum nor a relative minimum at =. (d) The graph of g has a point of inflection at each of =, =, and = because g ( ) = f ( ) changes sign at each of these values. : { : answer : eplanation The College Board. Visit the College Board on the Web:

4 SCORING GUIDELINES (Form B) Question 6 Let g be the piecewise-linear function defined on [ π, 4π ] whose graph is given above, and let f( ) = g( ) ( ) 4π cos. (a) Find f ( ) d. Show the computations that lead to your π answer. (b) Find all -values in the open interval ( π, 4π ) for which f has a critical point. (c) Let h ( ) gt ( ) π = Find h ( ). 4π 4π (a) ( ) = ( ( ) cos π ( )) π = 4 = π ( ) f d g d 6 sin = 6π π = π : { : antiderivative : answer (b) f ( ) g ( ) sin ( ) ( ) ( ) + sin for π < < = + = + sin for < < 4 π f ( ) does not eist at =. For π < <, f ( ). For < < 4 π, f ( ) = when = π. ( ( )) d : cos d 4 : : g ( ) : = : = π f has critical points at = and = π. (c) h ( ) = g( ) h π ( ) = g( π ) = π : h ( ) : : answer The College Board. Visit the College Board on the Web:

5 SCORING GUIDELINES Question 4 The continuous function f is defined on the interval 4. The graph of f consists of two quarter circles and one line segment, as shown in the figure above. Let g( ) = + f( t) (a) Find g(. ) Find g ( ) and evaluate g (. ) (b) Determine the -coordinate of the point at which g has an absolute maimum on the interval 4. Justify your answer. (c) Find all values of on the interval 4 < < 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. There is no point c, 4 < c <, for which f ( c) is equal to that average rate of change. Eplain why this statement does not contradict the Mean Value Theorem. 9π (a) g( ) = ( ) + f( t) dt = 6 4 g ( ) = + f( ) g ( ) = + f( ) = : : g( ) : g ( ) : g ( ) (b) g ( ) = when f( ) =. This occurs at = g 5 ( ) > for 4 < < and g ( ) < for 5 < <. 5 Therefore g has an absolute maimum at =. 5. : : considers g ( ) = : identifies interior candidate : answer with justification (c) g ( ) = f ( ) changes sign only at =. Thus the graph of g has a point of inflection at =. : answer with reason (d) The average rate of change of f on the interval 4 is f( ) f( 4) =. ( 4) 7 To apply the Mean Value Theorem, f must be differentiable at each point in the interval 4 < <. However, f is not differentiable at = and =. : average rate of change : { : eplanation The College Board. Visit the College Board on the Web:

6 9 SCORING GUIDELINES (Form B) Question A continuous function f is defined on the closed interval 4 6. The graph of f consists of a line segment and a curve that is tangent to the -ais at =, as shown in the figure above. On the interval < < 6, the function f is twice differentiable, with f ( ) >. (a) Is f differentiable at =? Use the definition of the derivative with one-sided limits to justify your answer. (b) For how many values of a, 4 a 6, equal to? Give a reason for your answer. (c) Is there a value of a, 4 a 6, < is the average rate of change of f on the interval [ a,6] < for which the Mean Value Theorem, applied to the interval [ a ] guarantees a value c, a < c < 6, at which f ( c) =? Justify your answer. (d) The function g is defined by g( ) = f( t) dt for 4 6. On what intervals contained in [ 4, 6] is the graph of g concave up? Eplain your reasoning. f( h) f( ) (a) lim = h h f( h) f( ) lim < + h h Since the one-sided limits do not agree, f is not differentiable at =. f( 6) f( a) (b) = when f( a) = f( 6. ) There are 6 a two values of a for which this is true. (c) Yes, a =. The function f is differentiable on the interval < < 6 and continuous on 6. f( 6) f( ) Also, = =. 6 6 By the Mean Value Theorem, there is a value c, < c < 6, such that f ( c) =. (d) g ( ) = f( ), g ( ) = f ( ) g ( ) > when f ( ) > This is true for 4 < < and < < 6.,6, : sets up difference quotient at = : { : answer with justification : epression for average rate of change : { : answer with reason : answers yes and identifies a = : { : justification : g ( ) = f( ) : : considers g ( ) > : answer 9 The College Board. All rights reserved. Visit the College Board on the Web:

7 The graph of the function f shown above consists of si line segments. Let g be the function given by g( ) = f( t) (a) Find g ( 4, ) g ( 4, ) and g ( 4. ) (b) Does g have a relative minimum, a relative maimum, or neither at =? Justify your answer. AP CALCULUS AB 6 SCORING GUIDELINES Question (c) Suppose that f is defined for all real numbers and is periodic with a period of length 5. The graph above shows two periods of f. Given that g ( 5) =, find g ( ) and write an equation for the line tangent to the graph of g at = 8. (a) g( 4) = f( t) dt = 4 g ( 4) = f( 4) = g ( 4) = f ( 4) = : : g( 4) : g ( 4) : g ( 4) (b) g has a relative minimum at = because g = f changes from negative to positive at =. : { : answer : reason (c) g ( ) = and the function values of g increase by for every increase of 5 in. g( ) = g( 5) = g( 8) = f() t dt + f() t dt 5 = g( 5) + g( ) = 44 4 : : g( ) : g( 8) : : g ( 8) : equation of tangent line g ( 8) = f( 8) = f( ) = An equation for the line tangent to the graph of g at = 8 is y 44 = ( 8 ). 6 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 4

8 5 SCORING GUIDELINES (Form B) Question 4 The graph of the function f above consists of three line segments. (a) Let g be the function given by g( ) = f( t) 4 For each of g(, ) g (, ) and g (, ) find the value or state that it does not eist. (b) For the function g defined in part (a), find the -coordinate of each point of inflection of the graph of g on the open interval 4 < <. Eplain your reasoning. (c) Let h be the function given by h ( ) = f( tdt ). Find all values of in the closed interval 4 for which h ( ) =. (d) For the function h defined in part (c), find all intervals on which h is decreasing. Eplain your reasoning. 5 (a) g( ) = f( t) dt = ( )( 5) = 4 g ( ) = f( ) = g ( ) does not eist because f is not differentiable at =. : : g( ) : g ( ) : g ( ) (b) = g = f changes from increasing to decreasing at =. : = (only) : : reason (c) =,, : correct values each missing or etra value (d) h is decreasing on [, ] h = f < when f > : interval : : reason Copyright 5 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 5

9 4 SCORING GUIDELINES Question 5 The graph of the function f shown above consists of a semicircle and three line segments. Let g be the function given by g( ) = f( t) (a) Find g ( ) and g (. ) (b) Find all values of in the open interval ( 5, 4) at which g attains a relative maimum. Justify your answer. (c) Find the absolute minimum value of g on the closed interval [ 5, 4 ]. Justify your answer. (d) Find all values of in the open interval ( 5, 4) at which the graph of g has a point of inflection. 9 (a) g( ) = f( t) dt = ( )( + ) = g ( ) = f( ) = : : g( ) : g ( ) (b) g has a relative maimum at =. This is the only -value where g = f changes from positive to negative. : = : : justification (c) The only -value where f changes from negative to positive is = 4. The other candidates for the location of the absolute minimum value are the endpoints. : identifies = 4 as a candidate : : g( 4) = : justification and answer g( 5) = 4 g( 4) = f( t) dt = g π ( 4) ( ) 9 π = + = So the absolute minimum value of g is. (d) =,, : correct values each missing or etra value Copyright 4 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6

10 SCORING GUIDELINES (Form B) Question 5 Let f be a function defined on the closed interval [,7]. The graph of f, consisting of four line segments, is shown above. Let g be the function given by g ( ) = ftdt ( ). (a) Find g (, ) g ( ), and g ( ). (b) Find the average rate of change of g on the interval. (c) For how many values c, where < c <, is g () c equal to the average rate found in part (b)? Eplain your reasoning. (d) Find the -coordinate of each point of inflection of the graph of g on the interval < < 7. Justify your answer. (a) g() = f( t) dt = ( 4 + ) = g () = f() = 4 g() = f() = = 4 : : g() : g() : g() (b) g() g() = () ftdt 7 ()(4) + (4 + ) = = ( ) : : g() g() = f( t) dt : answer (c) There are two values of c. We need 7 = g( c) = f( c) The graph of f intersects the line places between and. 7 y = at two : answer of : : reason Note: / if answer is by MVT (d) = and = 5 because g = f changes from increasing to decreasing at =, and from decreasing to increasing at = 5. : = and = 5 only : : justification (ignore discussion at = 4) Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com. 6

11 SCORING GUIDELINES (Form B) Question 4 6DACH=FDB=@EBBAHAJE=>ABK?JEBJDA?IA@ EJAHL=Г! # EIIDMEJDABECKHA=>LA6DACH=FDB BD=I=DHEJ=J=CAJEA=JN$AJ CN # BJ@J BH Г! > N > # $ N =.E@C$ C= $ =@C== $ > MD=JEJAHL=IEIC@A?HA=IECKIJEBOOKH=IMAH? MD=JEJAHL=IEIJDACH=FDBC??=LA@MKIJEBOOKH=IMAH BJ@J KIECIENIK>EJAHL=IBACJD J! Г! $ = C$ # # C$ $! C C = $ $ B$! C == $ C$ B$ > CEI@A?HA=IECГ! =@ # IE?A Г! C= N BN BHN =@N! # KIJEBE?=JE? 6DACH=FDBCEI??=LA@M$# IE?A EJAHL= C B EI@A?HA=IECJDEIEJAHL= KIJEBE?=JE! Г! Г Copyright by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 5

12 SCORING GUIDELINES Question 4 6DACH=FDBJDABK?JE B IDM=>LA?IEIJIBJMEAIACAJIAJ C >AJDA N BK?JECELA>O CN B =.E@ C Г C = Г =@ C == Г >.HMD=JL=KAIB N EJDAFAEJAHL= HA=IEC?.HMD=JL=KAIB N EJDAFAEJAHL= Г Г EI C E?HA=IEC-NF=EOKH JDA=NAIFHLE@A@IAJ?DJDACH=FDB C JDA?IA@EJAHL= Г = Г! CГ BJ@J Г BJ@J Г Г C= Г BГ C== Г B= Г! > CEIE?HA=IECГ N >A?=KIA C= N BN JDEIEJAHL=? 6DACH=FDBCEI??=LA@M N >A?=KIAC== N B= N JDEIEJAHL= H >A?=KIAC= N BN EI@A?HA=IECJDEI CГ! C= Г C== Г EJAHL= HA=I EJAHL= HA=I C Г C C =FFHFHE=JAE?HA=IEC@A?HA=IEC =@??=LEJO>AD=LEH Г LAHJE?==IOFJJA Copyright by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 5

13 AB{5 / BC{ The graph of the functionz f, consisting of three line segments, is given above. Let g() = f (t) (a) Compute g(4) and g(;). (b) Find the instantaneous rate of change of g, with respect to, at =. (c) Find the absolute minimum value of g on the closed interval [; 4]. Justify your answer. (d) The second derivative ofg is not dened at = and =. How many of these values are {coordinates of points of inection of the graph of g? Justify your answer. 4 (, 4) (, ) O 4 (4, ) Z 4 (a) g(4) = f (t) dt = ++ ; = 5 g(;) = Z ; f (t) dt = ; () = ;6 ( : g(4) : g(;) (b) g () = f () = 4 : answer (c) g is increasing on [; ] and decreasing on [ 4]. Therefore, g has absolute minimum at an endpoint of[; 4]. 8 : interior analysis >< : endpoint analysis >: : answer Since g(;) = ;6 and g(4) = 5, the absolute minimum value is ;6. (d) One = On (; ), g () =f () > On ( ), g () =f () < 8 : choice of = only >< : show (g()) is a point of inection >: : show (g()) is not a point of inection On ( 4), g () =f () < Therefore (g()) is a point of inection and (g()) is not.

AP CALCULUS AB 2001 SCORING GUIDELINES. Question 3

AP CALCULUS AB 2001 SCORING GUIDELINES. Question 3 AP CALCULUS AB SCORING GUIDELINES Question A car is traveling on a straight road with velocity ft/sec at time t =. For > t > 8 seconds, the car s acceleration at (), in ft/sec, is the piecewise linear