AP CALCULUS AB 2017 SCORING GUIDELINES

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1 AP CALCULUS AB 07 SCORING GUIDELINES

2 06 SCORING GUIDELINES Question 6 f ( ) f ( ) g( ) g ( ) The functions f and g have continuous second derivatives. The table above gives values of the functions and their derivatives at selected values of. (a) Let k( ) f ( g( ) ). Write an equation for the line tangent to the graph of k at. g( ) (b) Let h( ) f ( ). Find h (. ) (c) Evaluate f ( ) d. (a) k( ) f ( g( ) ) f ( 6) k ( ) f ( g( ) ) g ( ) f ( 6) 5 0 { : slope at : : equation for tangent line An equation for the tangent line is y 0( ) +. (b) h ( ) f ( ) g ( ) g( ) f ( ) ( f ( ) ) ( 6) 8 5 ( 6) 6 { : epression for : : answer h ( ) (c) f ( ) d ( ) [ ( 6) ( ) ] f f f 7 [ 5 ( ) ] : { : antiderivative : answer 06 The College Board. Visit the College Board on the Web:

3 Consider the curve given by the equation AP CALCULUS AB 05 SCORING GUIDELINES Question 6 dy y y. It can be shown that (a) Write an equation for the line tangent to the curve at the point (, ). y d y (b) Find the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical.. (c) Evaluate d y d at the point on the curve where and y. (a) dy d (, y) (, ) ( ) ( ) { : slope : : equation for tangent line An equation for the tangent line is y ( ). + + (b) (c) y 0 y So, y y y ( y )( y) y ( ) ( ) The tangent line to the curve is vertical at the point (, ). dy dy ( y ) y( 6y ) d y d d d y ( ) ( ) ( ) ( ) 6 d y d (, y) (, ) ( ( ) ) 6 : sets y 0 : : equation in one variable : coordinates : implicit differentiation dy : : substitution for d : answer 05 The College Board. Visit the College Board on the Web:

4 0 SCORING GUIDELINES Question 5 < < < < < < f( ) Positive 8 Positive Positive 7 f ( ) 5 Negative 0 Negative 0 Positive g( ) Negative 0 Positive Positive g ( ) Positive Positive 0 Negative The twice-differentiable functions f and g are defined for all real numbers. Values of f, f, g, and g for various values of are given in the table above. (a) Find the -coordinate of each relative minimum of f on the interval [, ]. Justify your answers. (b) Eplain why there must be a value c, for < c <, such that f ( c) 0. (c) The function h is defined by h ( ) ln ( f( ) ). Find h (. ) Show the computations that lead to your answer. (d) Evaluate f ( g( ) ) g ( ) d. (a) is the only critical point at which f changes sign from negative to positive. Therefore, f has a relative minimum at. (b) f is differentiable f is continuous on the interval f ( ) f ( ) ( ) : answer with justification : f ( ) f ( ) 0 : : eplanation, using Mean Value Theorem Therefore, by the Mean Value Theorem, there is at least one value c, < c <, such that f ( c) 0. (c) h ( ) f ( ) f( ) h ( ) f ( ) f ( ) 7 (d) f ( g( ) ) g ( ) d f ( g( ) ) f( g( ) ) f( g( ) ) f ( ) f ( ) 8 6 : h ( ) : : answer : { : Fundamental Theorem of Calculus : answer 0 The College Board. Visit the College Board on the Web:

5 0 SCORING GUIDELINES Question The function f is defined by f ( ) 5 for 5 5. (a) Find f ( ). (b) Write an equation for the line tangent to the graph of f at. f( ) for 5 (c) Let g be the function defined by g( ) + 7 for < 5. Is g continuous at? Use the definition of continuity to eplain your answer. (d) Find the value of d. f 5, 5 < < 5 : f ( ) 5 (a) ( ) ( ) ( ) (b) f ( ) 5 9 f ( ) 5 9 : f ( ) : : answer An equation for the tangent line is y + ( + ). (c) lim g( ) lim f( ) lim 5 lim g( ) lim ( + 7) + + Therefore, lim g( ). : considers one-sided limits : { : answer with eplanation g( ) f( ) So, lim g ( ) g( ). Therefore, g is continuous at. (d) Let u 5 du d d 0 u du 5 u 0 u u 5 5 ( 0 5) : { : antiderivative : answer 0 The College Board. Visit the College Board on the Web:

6 0 SCORING GUIDELINES (Form B) Question Consider a differentiable function f having domain all positive real numbers, and for which it is known that f ( ) ( ) for > 0. (a) Find the -coordinate of the critical point of f. Determine whether the point is a relative maimum, a relative minimum, or neither for the function f. Justify your answer. (b) Find all intervals on which the graph of f is concave down. Justify your answer. (c) Given that f ( ), determine the function f. (a) f ( ) 0 at f ( ) > 0 for 0 < < f ( ) < 0 for > Therefore f has a relative maimum at. : : : relative maimum : justification (b) f ( ) + ( )( ) + ( 6) ( 6) f ( ) < 0 for 0 < < 6 : f ( ) : : answer with justification The graph of f is concave down on the interval 0 < < 6. (c) f( ) + ( t t ) + t + t + dt t t : : integral : antiderivative : answer 0 The College Board. Visit the College Board on the Web:

7 0 SCORING GUIDELINES Question 6 sin for 0 Let f be a function defined by f( ) e for > 0. (a) Show that f is continuous at 0. (b) For 0, epress f ( ) as a piecewise-defined function. Find the value of for which f ( ). (c) Find the average value of f on the interval [, ]. (a) lim ( sin ) lim e f ( 0) So, lim f( ) f( 0 ). 0 : analysis Therefore f is continuous at 0. cos for 0 (b) f ( ) < e for > 0 cos for all values of < 0. e when ln ( ) > 0. Therefore f ( ) for ( ) ln. : f ( ) : : value of 0 (c) f( ) d f( ) d + f( ) d 0 0 ( sin ) d + e d cos + e ( cos( ) ) + ( e + ) 0 : 0 : ( sin ) d and e d 0 : antiderivatives : answer Average value ( ) f d cos( ) e The College Board. Visit the College Board on the Web:

8 The function g is defined for > 0 with (), AP CALCULUS AB 00 SCORING GUIDELINES (Form B) Question g g ( ) sin ( + ), and g ( ) ( + ) cos. (a) Find all values of in the interval 0. at which the graph of g has a horizontal tangent line. (b) On what subintervals of ( 0., ), if any, is the graph of g concave down? Justify your answer. (c) Write an equation for the line tangent to the graph of g at 0.. (d) Does the line tangent to the graph of g at 0. lie above or below the graph of g for 0. < <? Why? (a) The graph of g has a horizontal tangent line when g ( ) 0. This occurs at 0.6 and : sets g ( ) 0 : : answer (b) g ( ) 0 at and 0.7 The graph of g is concave down on ( 0.95, 0.7 ) because g ( ) < 0 on this interval. : { : answer : justification (c) g ( 0.) g( 0.) + g ( ) d An equation for the line tangent to the graph of g is y ( 0. ). : g ( 0.) : integral epression : : g ( 0. ) : equation (d) g ( ) > 0 for 0. < < : answer with reason Therefore the line tangent to the graph of g at 0. lies below the graph of g for 0. < <. 00 The College Board. Visit the College Board on the Web:

9 008 SCORING GUIDELINES (Form B) Consider the closed curve in the y-plane given by (a) Show that dy d ( + ). y + ( ) Question y + y 5. (b) Write an equation for the line tangent to the curve at the point (, ). (c) Find the coordinates of the two points on the curve where the line tangent to the curve is vertical. (d) Is it possible for this curve to have a horizontal tangent at points where it intersects the -ais? Eplain your reasoning. (a) dy dy + + y + 0 d d dy ( y + ) d dy ( + ) ( + ) d y + y + ( ) ( ) : implicit differentiation : { : verification (b) ( ) dy + d (, ) ( + ) y + + Tangent line: ( ) : { : slope : tangent line equation (c) Vertical tangent lines occur at points on the curve where y + 0 (or y ) and. On the curve, y implies that so or , : : y : substitutes y into the equation of the curve : answer Vertical tangent lines occur at the points (, ) and (, ). (d) Horizontal tangents occur at points on the curve where and y. : works with or y 0 : { : answer with reason The curve crosses the -ais where y 0. ( ) + ( ) No, the curve cannot have a horizontal tangent where it crosses the -ais. 008 The College Board. All rights reserved. Visit the College Board on the Web:

10 008 SCORING GUIDELINES (Form B) Question The functions f and g are given by f ( ) + t dt and g( ) f( sin ). 0 (a) Find f ( ) and g ( ). (b) Write an equation for the line tangent to the graph of y g( ) at π. (c) Write, but do not evaluate, an integral epression that represents the maimum value of g on the interval 0 π. Justify your answer. (a) f ( ) + ( ) g ( ) f ( sin ) cos : : : f ( ) g ( ) + ( sin ) cos (b) g( π ) 0, g ( π ) 6 Tangent line: y 6( π ) : g( π) or g ( π) : : tangent line equation (c) For 0 < < π, g π ( ) 0 only at. g( 0) g( π ) 0 ( ) π + > 0 g t dt 0 0, π is The maimum value of g on [ ] 0 + t dt. : : sets g ( ) 0 π : justifies maimum at : integral epression for g π ( ) 008 The College Board. All rights reserved. Visit the College Board on the Web:

11 008 SCORING GUIDELINES Question 6 ln Let f be the function given by f( ) for all > 0. The derivative of f is given by ln f ( ). (a) Write an equation for the line tangent to the graph of f at e. (b) Find the -coordinate of the critical point of f. Determine whether this point is a relative minimum, a relative maimum, or neither for the function f. Justify your answer. (c) The graph of the function f has eactly one point of inflection. Find the -coordinate of this point. (d) Find lim f ( ). + 0 ln (a) f( e ) e, lne f ( e ) e e ( e ) e An equation for the tangent line is y ( e ) e e. ( ) ( ) : : f e and f e : answer (b) f ( ) 0 when e. The function f has a relative maimum at e because f ( ) changes from positive to negative at e. : : e : relative maimum : justification ( ln ) ln (c) f ( ) + for all > 0 f ( ) 0 when + ln 0 : f ( ) : : answer e The graph of f has a point of inflection at f ( ) changes sign at e. e because (d) ln lim + 0 or Does Not Eist : answer 008 The College Board. All rights reserved. Visit the College Board on the Web:

12 007 SCORING GUIDELINES (Form B) Question 6 Let f be a twice-differentiable function such that f ( ) 5 and f ( 5). Let g be the function given by g( ) f ( f ( ) ). (a) Eplain why there must be a value c for < c < 5 such that f ( c ). (b) Show that g ( ) g ( 5 ). Use this result to eplain why there must be a value k for < k < 5 such that g ( k ) 0. (c) Show that if f ( ) 0 for all, then the graph of g does not have a point of inflection. (d) Let h ( ) f( ). Eplain why there must be a value r for < r < 5 such that hr ( ) 0. (a) The Mean Value Theorem guarantees that there is a value c, with < c < 5, so that f( 5) f( ) 5 f ( c). 5 5 f( 5) f( ) : : 5 : conclusion, using MVT (b) g ( ) f ( f( ) ) f ( ) g ( ) f ( f ( ) ) f ( ) f ( 5) f ( ) g ( 5) f ( f( 5) ) f ( 5) f ( ) f ( 5 ) Thus, g ( ) g ( 5). Since f is twice-differentiable, g is differentiable everywhere, so the Mean Value Theorem applied to g on [, 5] guarantees there is a value k, with < k < 5, such that g ( k) g ( 5) g ( ) 0. 5 : : g ( ) : g ( ) f ( 5) f ( ) g ( 5) : uses MVT with g (c) g ( ) f ( f( ) ) f ( ) f ( ) + f ( f( ) ) f ( ) If f ( ) 0 for all, then g ( ) 0 f ( ) f ( ) + f ( f( ) ) 0 0 for all. Thus, there is no -value at which g ( ) changes sign, so the graph of g has no inflection points. OR If f ( ) 0 for all, then f is linear, so g f f is linear and the graph of g has no inflection points. (d) Let h( ) f( ). h( ) f( ) 5 h( 5) f( 5) 5 5 Since h( ) > 0 > h( 5), the Intermediate Value Theorem guarantees that there is a value r, with < r < 5, such that h( r ) 0. : considers g : : g ( ) 0 for all OR : f is linear : { : g is linear : h( ) and h( 5 ) : : conclusion, using IVT 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

13 007 SCORING GUIDELINES Question 6 Let f be the function defined by f ( ) k ln for > 0, where k is a positive constant. (a) Find f ( ) and f ( ). (b) For what value of the constant k does f have a critical point at? For this value of k, determine whether f has a relative minimum, relative maimum, or neither at. Justify your answer. (c) For a certain value of the constant k, the graph of f has a point of inflection on the -ais. Find this value of k. (a) k f ( ) : : : f ( ) f ( ) f ( ) k + (b) f () k 0 k When k, f () 0 and f () + > 0. f has a relative minimum value at by the Second Derivative Test. : : sets f () 0 or f ( ) 0 : solves for k : answer : justification (c) At this inflection point, f ( ) 0 and f ( ) 0. k f ( ) k ln f( ) 0 k ln 0 k : : f ( ) 0 or f( ) 0 : equation in one variable : answer Therefore, ln ln e k e 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

14 007 SCORING GUIDELINES Question f ( ) f ( ) g( ) g ( ) 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. The function h is given by h ( ) f( g ( )) 6. (a) Eplain why there must be a value r for < r < such that hr ( ) 5. (b) Eplain why there must be a value c for < c < such that h ( c) 5. g( ) (c) Let w be the function given by w ( ) f( t) dt. Find the value of w (. ) (d) If g is the inverse function of g, write an equation for the line tangent to the graph of y g ( ) at. (a) h() f( g() ) 6 f( ) h( ) f( g( ) ) 6 f( ) Since h( ) < 5 < h( ) and h is continuous, by the Intermediate Value Theorem, there eists a value r, < r <, such that h( r ) 5. (b) h( ) h( ) 7 5 Since h is continuous and differentiable, by the Mean Value Theorem, there eists a value c, < c <, such that h ( c) 5. (c) w ( ) f ( g( ) ) g ( ) f ( ) (d) g (), so g ( ). ( g ) ( ) g g ( ) g 5 ( ) () An equation of the tangent line is y ( ). 5 : h() and h( ) : : conclusion, using IVT h( ) h( ) : : : conclusion, using MVT : apply chain rule : { : answer : g ( ) : : ( g ) ( ) : tangent line equation 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

15 006 SCORING GUIDELINES (Form B) Question The figure above is the graph of a function of, which models the height of a skateboard ramp. The function meets the following requirements. (i) At 0, the value of the function is 0, and the slope of the graph of the function is 0. (ii) At, the value of the function is, and the slope of the graph of the function is. (iii) Between 0 and, the function is increasing. (a) Let f ( ) a, where a is a nonzero constant. Show that it is not possible to find a value for a so that f meets requirement (ii) above. (b) Let g( ) c, where c is a nonzero constant. Find the value of c so that g meets requirement (ii) 6 above. Show the work that leads to your answer. (c) Using the function g and your value of c from part (b), show that g does not meet requirement (iii) above. n (d) Let h ( ), where k is a nonzero constant and n is a positive integer. Find the values of k and n so that k h meets requirement (ii) above. Show that h also meets requirements (i) and (iii) above. (a) f ( ) implies that implies that a and f ( ) a( ) 6 a. Thus, f cannot satisfy (ii). 8 : a or a : 6 8 : shows a does not work (b) g( ) 6c implies that c. When, ( ) c g ( ) c( ) ( )( 6) 6 (c) g ( ) ( ) 8 g ( ) < 0 for 0 < <, so g does not satisfy (iii). n n (d) h( ) implies that k. k n n n n n h ( ) gives n and k n h( ) h( 0) h ( ) h ( 0) 0 and h ( ) > 0 for 0 < <. 56 k 56. : value of c : g ( ) : : eplanation n : k n- : n : k : values for k and n : verifications 006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents).

16 006 SCORING GUIDELINES Question 6 The twice-differentiable function f is defined for all real numbers and satisfies the following conditions: f ( 0), f ( 0), and f ( 0). a (a) The function g is given by g ( ) e + f ( ) for all real numbers, where a is a constant. Find g ( 0) and g ( 0) in terms of a. Show the work that leads to your answers. (b) The function h is given by h ( ) cos( k) f( ) for all real numbers, where k is a constant. Find h ( ) and write an equation for the line tangent to the graph of h at 0. (a) g ( ) ae a + f ( ) g ( 0) a a g ( ) a e + f ( ) g ( 0) a + : : g ( ) : g ( 0) : g ( ) : g ( 0) (b) h ( ) f ( ) cos( k) ksin ( k) f( ) h ( 0) f ( 0) cos( 0) ksin( 0) f( 0) f ( 0) h( 0) cos( 0) f( 0) The equation of the tangent line is y +. 5 : : h ( ) : h ( 0) : : h( 0) : equation of tangent line 006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 7

17 005 SCORING GUIDELINES (Form B) Question 5 Consider the curve given by y + y. dy y (a) Show that. d y (b) Find all points (, y ) on the curve where the line tangent to the curve has slope. (c) Show that there are no points (, y ) on the curve where the line tangent to the curve is horizontal. (d) Let and y be functions of time t that are related by the equation dy value of y is and 6. Find the value of d at time t 5. dt dt y + y. At time t 5, the (a) yy y + y ( y ) y y y y y : implicit differentiation : : solves for y (b) y y y y 0 y ± ( 0, ), ( 0, ) y : : y : answer y (c) 0 y y 0 The curve has no horizontal tangent since for any. : y 0 : : eplanation (d) When y, dt t 5 + so dy dy d y d dt d dt y dt d 9 d At t 5, 6 7 dt dt 6 d 7. : : solves for : chain rule : answer Copyright 005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6

18 005 SCORING GUIDELINES Question 0 0 < < < < < < < < f ( ) Negative 0 Positive Positive 0 Negative f ( ) Positive 0 Positive DNE Negative Negative f ( ) Negative 0 Positive DNE Negative 0 Positive 0,. The function f is twice differentiable ecept at. The function f and its derivatives have the properties indicated in the table above, where DNE indicates that the derivatives of f do not eist at. (a) For 0 < <, find all values of at which f has a relative etremum. Determine whether f has a relative maimum or a relative minimum at each of these values. Justify your answer. (b) On the aes provided, sketch the graph of a function that has all the characteristics of f. (Note: Use the aes provided in the pink test booklet.) Let f be a function that is continuous on the interval [ ) (c) Let g be the function defined by g( ) f( t) dt on the open interval ( 0, ). For 0 < <, find all values of at which g has a relative etremum. Determine whether g has a relative maimum 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, for 0 < <, at which the graph of g has a point of inflection. Justify your answer. (a) f has a relative maimum at because f changes from : relative etremum at : positive to negative at. { : relative maimum with justification (b) : : points at 0,,, and behavior at (, ) : appropriate increasing/decreasing and concavity behavior (c) g ( ) f ( ) 0 at,. g changes from negative to positive at so g has a relative minimum at. g changes from positive to negative at so g has a relative maimum at. (d) The graph of g has a point of inflection at because g f changes sign at. : : g ( ) f( ) : critical points : answer with justification : { : : answer with justification Copyright 005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 5

19 00 SCORING GUIDELINES Question Consider the curve given by + y 7 + y. dy y (a) Show that. d 8y (b) Show that there is a point P with -coordinate at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P. d y (c) Find the value of at the point P found in part (b). Does the curve have a local maimum, a d local minimum, or neither at the point P? Justify your answer. (a) + 8yy y + y ( 8y ) y y y y 8y : implicit differentiation : : solves for y (b) y 8y 0; y 0 When, y 6 y + 5 and Therefore, P (, ) is on the curve and the slope is 0 at this point. dy : 0 d : : shows slope is 0 at (, ) : shows (, ) lies on curve (c) d y ( 8y )( y ) ( y )( 8y ) d ( 8y ) d y ( 6 9)( ) At P (, ),. d ( 6 9) 7 Since y 0 and y < 0 at P, the curve has a local maimum at P. : d y : d d y : value of at (, ) d : conclusion with justification Copyright 00 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 5

20 00 SCORING GUIDELINES Question 6 Let f be the function defined by + for 0 f () 5 for < 5. (a) Is f continuous at? Eplain why or why not. (b) Find the average value of f () on the closed interval 0 5. (c) Suppose the function g is defined by k + for 0 g () m + for < 5, where k and m are constants. If g is differentiable at, what are the values of k and m? (a) f is continuous at because lim f ( ) lim f( ). + Therefore, lim f ( ) f(). : : answers yes and equates the values of the left- and right-hand limits : eplanation involving limits (b) 5 5 f ( ) d f ( ) d + f ( ) d 0 0 Average value: ( ) ( 5 ) ( ) ( ) ( ) 5 fd 0 5 : 5 : k f( ) d + k ( ) 0 f d (where k 0) : antiderivative of + : antiderivative of 5 : evaluation and answer (c) Since g is continuous at, k m +. k for 0 < < g ( ) + m for < < 5 k lim g ( ) and lim g ( ) m + Since these two limits eist and g is differentiable at, the two limits are k equal. Thus m. : k m + : k : m : values for k and m 8 8m m + ; m and k 5 5 Copyright 00 by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com. 7

21 00 SCORING GUIDELINES N BN Г# Г Question 6 Г# # # Г Г" Г$ Г% Г$ Г" Г B N Г% Г# Г!! # % AJ B B BHIAA?JA@FEJI N EJDA?IA@EJAHLГ # > N > # 6DAIA?@@AHELJELAB B DIJDA FHFAHJOJDJ B N BHГ # > N > # # -LKJA! B N > 9HEJAAGKJEBJDAEAJCAJJJDACHFDB B JJDAFEJMDAHA N 7IAJDEIEAJ FFHNEJAJDALKAB B IJDEIFFHNEJECHAJAHJDHAIIJDJDA?JKLKAB B /ELAHAIBHOKHIMAH?.E@FIEJELAHAK>AH H DLECJDAFHFAHJOJDJJDAHAKIJANEIJLKA? MEJD? B? H/ELAHAIBHOKHIMAH N ГN Г% BH AJ C >AJDABK?JECELA>O CN N N Г% BH N F 6DACHFDB C FIIAIJDHKCDA?DBJDAFEJI NBN C HAJDAIABK?JE/ELAHAIBHOKHIMAH # # #! B N B "@N #! BN " N! ГГ Г% " # " > O # N Г Г " B N # Г " Г! 6DAFFHNEJEEIAIIJD B >A?KIAJDA CHFDBBEI??LAKFJDAEJAHL N? *OJDAA8KA6DAHAJDAHAEI?MEJD? # IK?DJDJ B # Г B! Г B? $ H # Г E C N E " N Г Г N Г N Г C N N N N E E " 6DKI C EIJ?JEKKIJN>KJ B EI?JEKKIJNI B L C C N " BH N L >KJEJMIIDMEFHJ?JDJ B? $ BHIA? L I B L C JE@AHELJELA IMAH JCAJEA!?FKJAIOJCAJEAJN IMAHMEJDHAI HABAHA?AJ86BHB H@EBBAHAJE>EEJO BB LKABHBHEJAHL# > N > IMAHI]^MEJDHABAHA?AJ C HC?HHA?JHAI Copyright 00 by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 7

22 00 SCORING GUIDELINES Question )>A?JLAICJDA NNEIMEJDEEJEFIEJE N 6DALA?EJOBJDA>A?JJJEAJ F EICELA >O L J IE J! 9DJEIJDA??AAHJEBJDA>A?JJJEA J " > +IE@AHJDABMECJMIJJAAJI 5JJAAJ.H! J "#JDALA?EJOBJDA>A?JEI@A?HAIEC 5JJAAJ.H! J "#JDAIFAA@BJDA>A?JEIE?HAIEC )HAAEJDAHH>JDBJDAIAIJJAAJI?HHA?J.HA?DIJJAAJFHLE@AHAIMDOEJEI?HHA?JHJ?HHA?J? 9DJEIJDAJJ@EIJ?AJHLAA@>OJDA>A?JLAHJDAJEAEJAHL > J > 9DJEIJDAFIEJEBJDA>A?JJJEA J " " " L "?I!! Г HГ#! HГ# " $ IMAH >! J "# J L J?I J!! 5JJAAJEI?HHA?JIE?AJ J?,EIJ?A " NJ Г?I J! N N" '!"!!' LJ MDAJ! $ N!!''&$ N!ГN N"ГN! N" N NJ Г?I J! N" '!"! " Copyright 00 by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board.?HHA?JMEJDHAI!?HHA?J HAIBH EEJIB@"EJACH BLJ H LJ H D@AI?DCAB@EHA?JEJ IJK@AJ\IJKHECFEJ IMAH EBE?HHA?JJKHECFEJH JKHECFEJ EJACH IMAH! NJ Г?I J +! IMAH EB?IJJBEJACHJE

23 AP CALCULUS AB 00 SCORING GUIDELINES Question Let h be a function defined for all L 0 such that h() Г and the derivative of h is given Г by h ( ) for all L 0. (a) Find all values of for which the graph of h has a horizontal tangent, and determine whether h has a local maimum, a local minimum, or neither at each of these values. Justify your answers. (b) On what intervals, if any, is the graph of h concave up? Justify your answer. (c) Write an equation for the line tangent to the graph of h at. (d) Does the line tangent to the graph of h at lie above or below the graph of h for? Why? (a) h ( ) 0 at h ( ) 0 + und 0 + Г 0 : : : analysis : conclusions Г > not dealing with discontinuity at 0 Local minima at Г and at (b) h ( ) 0 for all L 0. Therefore, the graph of h is concave up for all L 0. : : h ( ) : h ( ) 0 : answer (c) 6 Г 7 h () 7 y ( Г ) : tangent line equation (d) The tangent line is below the graph because the graph of h is concave up for. : answer with reason Copyright 00 by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 5

24 !0Å#ALCULUSÅ!"nÅÅ"#n #ONSIDERÅTHEÅCURVEÅGIVENÅBYÅXY X Y A DY X Y Y HOWÅTHATÅ DX XY X B &INDÅALLÅPOINTSÅONÅTHEÅCURVEÅWHOSEÅXCOORDINATEÅISÅÅANDÅWRITEÅANÅEQUATIONÅFORÅTHEÅTANGENTÅLINEÅAT EACHÅOFÅTHESEÅPOINTS C &INDÅTHEÅXCOORDINATEÅOFÅEACHÅPOINTÅONÅTHEÅCURVEÅWHEREÅTHEÅTANGENTÅLINEÅISÅVERTICAL A DY DY Y XY X Y X DX DX DY DX XY X X Y Y IMPLICITÅDIFFERENTIATION ÅÅVERIFIESÅEXPRESSIONÅFORÅ DY DX DY X Y Y DX XY X B 7HENÅXÅÅ Y Y Y Y Y Y YÅÅÅYÅÅn DY!TÅ Å DX ANGENTÅLINEÅEQUATIONÅISÅYÅÅ ÅY Y Å ÅSOLVESÅFORÅY ÅTANGENTÅLINESÅ.OTEÅÅIFÅNOTÅSOLVINGÅANÅEQUATIONÅOFÅTHE FORMÅY Y K DY!TÅ Å DX ANGENTÅLINEÅEQUATIONÅISÅY X C ANGENTÅLINEÅISÅVERTICALÅWHENÅXY X X Y X ÅGIVESÅXÅÅÅORÅ Y X HEREÅISÅNOÅPOINTÅONÅTHEÅCURVEÅWITH XCOORDINATEÅ DY ÅSETSÅDENOMINATORÅOFÅ ÅEQUALÅTOÅ DX ÅSUBSTITUTESÅY X ÅORÅX o YÅ Å ÅÅÅÅÅÅINTOÅTHEÅEQUATIONÅFORÅTHEÅCURVE ÅSOLVESÅFORÅX COORDINATE 7HENÅY X ÅÅ X ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ X X ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅX Copyright 000 by College Entrance Eamination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Eamination Board.

25 AB{ 999. Suppose that the function f has a continuous second derivative for all, and that f(0), f 0 (0) ;, and f 00 (0) 0. Let g be a function whose derivative isgiven by g 0 () e ; (f()+f 0 ()) for all. (a) Write an equation of the line tangent to the graph of f at the point where 0. (b) Is there sucient information to determine whether or not the graph of f has a point of inection when 0? Eplain your answer. (c) Given that g(0), write an equation of the line tangent to the graph of g at the point where 0. (d) Show that g 00 () e ; (;6f() ; f 0 ()+f 00 ()). Does g have a local maimum at 0? Justify your answer. (a) Slope at 0isf 0 (0) ; At 0,y y ; ;( ; 0) : equation (b) No. Whether f 00 () changes sign at 0is unknown. The only given value of f 00 () is f 00 (0) 0. ( : answer : eplanation (c) g 0 () e ; (f()+f 0 ()) g 0 (0) e 0 (f(0)+f 0 (0)) ( : g0 (0) : equation ()+(;) 0 y ; 0( ; 0) y (d) g 0 () e ; (f()+f 0 ()) g 00 () (;e ; )(f()+f 0 ()) + e ; (f 0 ()+f 00 ()) e ; (;6f() ; f 0 ()+f 00 ()) 8 >< >: : verify derivative 0/ product or chain rule error <; > algebra errors : g 0 (0) 0 and g 00 (0) : answer and reasoning g 00 (0) e 0 [(;6)() ; (;) + (0)] ;9 Since g 0 (0) 0 and g 00 (0) < 0, g does have a local maimum at 0.

26 998 AP Calculus AB Scoring Guidelines 6. Consider the curve defined by y + 6 y + 6y. (a) Show that dy d y + y +. (b) Write an equation of each horizontal tangent line to the curve. (c) The line through the origin with slope is tangent to the curve at point P. Find the and y coordinates of point P. (a) 6y dy dy dy y + 6 d d d 0 dy d (6y ) y dy d y 6 + 6y + 6 y + y + : implicit differentiation : verifies epression for dy d (b) (c) dy d 0 y ( y) 0 0 or y When 0, y + 6y ; y 0.65 There is no point on the curve with y coordinate of. y 0.65 is the equation of the only horizontal tangent line. y is equation of the line. ( ) + 6 ( ) + 6( ) /, y / or dy d y y /, y / : sets dy d 0 : solves dy d 0 : uses solutions for to find equations of horizontal tangent lines : verifies which solutions for y yield equations of horizontal tangent lines Note: ma / [-0-0-0] if dy/d 0 is not of the form g(, y)/h(, y) 0 with solutions for both and y : y : substitutes y into equation of curve : solves for and y or : sets dy d : substitutes y into dy d : solves for and y Note: ma / [--0] if importing incorrect derivative from part (a) Copyright 998 College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board.

27 . Let f be the function given by f() e. 998 Calculus AB Scoring Guidelines (a) Find lim f() and lim f(). (b) Find the absolute minimum value of f. Justify that your answer is an absolute minimum. (c) What is the range of f? (d) Consider the family of functions defined by y be b, where b is a nonzero constant. Show that the absolute minimum value of be b is the same for all nonzero values of b. (a) lim e 0 lim e or DNE { : 0 as : or DNE as (b) f () e + e e ( + ) 0 if / f( /) /e or 0.68 or 0.67 /e is an absolute minimum value because: (i) (ii) f () < 0 for all < / and f () > 0 for all > / f () or + / and / is the only critical number : solves f () 0 : evaluates f at student s critical point 0/ if not local minimum from student s derivative : justifies absolute minimum value 0/ for a local argument 0/ without eplicit symbolic derivative Note: 0/ if no absolute minimum based on student s derivative (c) Range of f [ /e, ) or [ 0.67, ) or [ 0.68, ) (d) y be b + b e b be b ( + b) 0 if /b At /b, y /e y has an absolute minimum value of /e for all nonzero b : answer Note: must include the left hand endpoint; eclude the right hand endpoint : sets y be b ( + b) 0 : solves student s y 0 : evaluates y at a critical number and gets a value independent of b Note: 0/ if only considering specific values of b Copyright 998 College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board.