05 SCORING GUIDELINES Question 5 Consider the function f =, where k is a nonzero constant. The derivative of f is given by k f = k ( k). (a) Let k =, so that f =. Write an equation for the line tangent to the graph of f at the point whose -coordinate is 4. (b) Let k = 4, so that f =. Determine whether f has a relative minimum, a relative maimum, or 4 neither at =. Justify your answer. (c) Find the value of k for which f has a critical point at = 5. (d) Let k = 6, so that f =. Find the partial fraction decomposition for the function f. 6 Find f d. 4 5 (a) f( 4) = = f (4) = = 4 4 4 6 4 4 An equation for the line tangent to the graph of f at the point 5 whose -coordinate is 4 is y = ( 4 ) +. 6 4 4 4 (b) f = f = = 0 4 4 f changes sign from positive to negative at =. Therefore, f has a relative maimum at =. { : slope : : tangent line equation : considers f : : answer with justification k ( 5) (c) f ( 5) = = 0 k = 0 ( 5) k ( 5) (d) A B A( 6) B 6 = ( 6) = + 6 = + = 0 = A ( 6) A = 6 = 6 = B ( 6) B = 6 6 6 = + ( 6) 6 6 6 f d = + d 6 6 = ln + ln 6 + C = ln + C 6 6 6 : answer 05 The College Board. Visit the College Board on the Web: www.collegeboard.org. { : partial fraction decomposition 4: : general antiderivative
0 SCORING GUIDELINES (Form B) Question 4 The graph of the differentiable function y = f with domain 0 0 is shown in the figure above. The area of the region enclosed between the graph of f and the -ais for 0 5 is 0, and the area of the region enclosed between the graph of f and the -ais for 5 0 is 7. The arc length for the portion of the graph of f between = 0 and = 5 is, and the arc length for the portion of the graph of f between = 5 and = 0 is 8. The function f has eactly two critical points that are located at = and = 8. (a) Find the average value of f on the interval 0 5. 0 (b) Evaluate ( f + ) d. Show the computations that lead to 0 your answer. 5 (c) Let g = f( t) dt. On what intervals, if any, is the graph of g both concave up and decreasing? Eplain your reasoning. (d) The function h is defined by h = f. The derivative of h is h f. the graph of y = h from = 0 to = 0. = Find the arc length of 5 0 (a) Average value = f d 5 = = : answer 0 5 0 5 0 (b) ( f + ) d = f d + f d + 0 0 0 5 = ( 0 + 7) + 0 = 7 : answer (c) g = f g < 0 on 0 < < 5 g is increasing on < < 8. The graph of g is concave up and decreasing on < < 5. : : g = f : analysis : answer and reason 0 ( ) 0 (d) Arc length = + h d = + f d 0 0 Let u =. Then du = d and 0 ( ) + f d = + ( f u ) du = ( + 8) = 58 0 0 0 : : integral : substitution : answer 0 The College Board. Visit the College Board on the Web: www.collegeboard.org.
00 SCORING GUIDELINES (Form B) Question 5 4 Let f and g be the functions defined by f = and g =, for all > 0. + 4 (a) Find the absolute maimum value of g on the open interval ( 0, ) if the maimum eists. Find the absolute minimum value of g on the open interval ( 0, ) if the minimum eists. Justify your answers. (b) Find the area of the unbounded region in the first quadrant to the right of the vertical line =, below the graph of f, and above the graph of g. 4+ 4 4 8 4 4 (a) g = = ( + 4 ) + 4 For > 0, g = 0 for =. g > 0 for 0 < < g < 0 for > g = 5 : : g : critical point : answers : justification Therefore g has a maimum value of at =, and g has no minimum value on the open interval ( 0, ). = lim ( ln ln ( + 4 )) (b) ( f g ) d = lim ( f g ) d b b = b = b ( ( b) ( b ) ) = lim ln ln + 4 + ln 5 b b 5 = lim ln b + 4b 5b = lim ln b + 4b 4 : : integral : antidifferentiation : answer 5b = lim ln b + 4b = 5 ln 4 00 The College Board. Visit the College Board on the Web: www.collegeboard.com.
009 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 0 < < 6, the function f is twice differentiable, with f > 0. (a) Is f differentiable at = 0? 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 0? 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( 0) (a) lim = h 0 h f( h) f( 0) lim < 0 + h 0 h Since the one-sided limits do not agree, f is not differentiable at = 0. f( 6) f( a) (b) = 0 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 0 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 > 0 when f > 0 This is true for 4 < < 0 and < < 6. 0,6, : sets up difference quotient at = 0 : { : answer with justification : epression for average rate of change : { : answer with reason : answers yes and identifies a = : { : justification : g = f : : considers g > 0 : answer 009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.
008 SCORING GUIDELINES Question 5 The derivative of a function f is given by f = ( ) e for > 0, and f () = 7. (a) The function f has a critical point at =. At this point, does f have a relative minimum, a relative maimum, or neither? Justify your answer. (b) On what intervals, if any, is the graph of f both decreasing and concave up? Eplain your reasoning. (c) Find the value of f (. ) (a) f < 0 for 0 < < and f > 0 for > Therefore, f has a relative minimum at =. : minimum at = : : justification (b) f = e + ( ) e = ( ) e f > 0 for > : f : : answer with reason f < 0 for 0 < < Therefore, the graph of f is both decreasing and concave up on the interval < <. (c) f = f + f d = 7 + ( ) e d u = dv = e d du = d v = e f = 7 + ( ) e e d = 7 + (( ) e e ) = 7 + e e 4: : uses initial condition : integration by parts : answer 008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.
007 SCORING GUIDELINES Question 4 Let f be the function defined for > 0, with f( e ) = and f, the first derivative of f, given by f = ln. (a) Write an equation for the line tangent to the graph of f at the point ( e,). (b) Is the graph of f concave up or concave down on the interval < <? Give a reason for your answer. (c) Use antidifferentiation to find f ( ). (a) f ( e) = e An equation for the line tangent to the graph of f at the point ( e,) is y = e ( e). : f ( e) : : equation of tangent line (b) f = + ln. For < <, > 0 and ln > 0, so f > 0. Thus, the graph of f is concave up on (, ). : f : : answer with reason f = ln d, we consider integration by (c) Since parts. u = ln dv = d du = d v = ( ) d = : antiderivative 4 : : uses f( e) = : answer Therefore, = ( ln ) f d = ln d = ln + C. 9 e e Since f( e ) =, = + C and C 9 Thus, f = ln + e. 9 9 = e. 9 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).
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 = 4, the value of the function is, and the slope of the graph of the function is. (iii) Between = 0 and = 4, 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 ( 4) = implies that implies that a = and f ( 4) = a( 4) = 6 a =. Thus, f cannot satisfy (ii). 8 : a = or a = : 6 8 : shows a does not work (b) g( 4) = 64c = implies that c =. When, 4 c = g ( 4) = c( 4) = ( 6) = 6 (c) g = = ( 4) 8 g 4 < 0 for 0 < <, so g does not satisfy (iii). n 4 n (d) h( 4) = = implies that 4 = k. k n n n4 n4 n h ( 4) = = = = gives n = 4 and k n 4 4 4 h = h( 0) = 0. 56 4 h = h ( 0) = 0 and h > 0 for 0 < < 4. 56 4 k = 4 = 56. : value of c : g : : eplanation n 4 : = k n- 4 : n4 : = k : values for k and n : verifications 006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 4
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 0 + 0 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 www.collegeboard.com/apstudents (for AP students and parents). 6
004 SCORING GUIDELINES Question 4 Consider the curve given by + 4y = 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 = + 4 = 5 and 7 + = 5 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. 4 : d y : d d y : value of at (, ) d : conclusion with justification Copyright 004 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 5
AP CALCULUS BC 00 SCORING GUIDELINES Question 4 Let h be a function defined for all L 0 such that h(4) 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 = 4. (d) Does the line tangent to the graph of h at = 4 lie above or below the graph of h for 4? Why? (a) h= 0 at h= 0 + und 0 + Г 0 4 : : : 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= (4) 4 7 y ( Г 4) : tangent line equation (d) The tangent line is below the graph because the graph of h is concave up for 4. : 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
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998 Calculus BC Scoring Guidelines Copyright 998 by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board.