AP CALCULUS AB 2011 SCORING GUIDELINES (Form B)

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1 0 SCORING GUIDELINES (Form B) Consider a differentiable function f having domain all positive real numbers, and for which it is known that f = ( 4 x) x for x > 0. (a) Find the x-coordinate of the critical point of f. Determine whether the point is a relative maximum, 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 x = 4 f > 0 for 0 < x < 4 f < 0 for x > 4 Therefore f has a relative maximum at x = 4. : x = 4 : relative maximum : justification 4 (b) f = x + ( 4 x)( x ) 4 = x x + x 4 = x ( x 6) ( x 6) = 4 x f < 0 for 0 < x < 6 : f : answer with justification The graph of f is concave down on the interval 0 < x < 6. x (c) f = + ( 4t t ) = + t + t = x + x dt t= x t= : integral : antiderivative : answer 0 The College Board. Visit the College Board on the Web:

2 009 SCORING GUIDELINES (Form B) Question Let f be a twice-differentiable function defined on the interval. < x <. with f () =. The graph of f, the derivative of f, is shown above. The graph of f crosses the x-axis at x = and x = and has a horizontal tangent at x =. Let g f be the function given by g = e. (a) Write an equation for the line tangent to the graph of g at x =. (b) For. < x <., find all values of x at which g has a local maximum. Justify your answer. f (c) The second derivative of g is g = e ( f ) + f. g ( ) positive, negative, or zero? Justify your answer. (d) Find the average rate of change of g, the derivative of g, over the interval [, ]. f () (a) g() = e = e f f () g = e f, g () = e f () = 4e The tangent line is given by y = e 4e ( x ). : g : g() and g () : tangent line equation f (b) g = e f f e > 0 for all x So, g changes from positive to negative only when f changes from positive to negative. This occurs at x = only. Thus, g has a local maximum at x =. : { : answer : justification ( ) f (c) g ( ) = e ( f ( ) ) + f ( ) f ( ) e > 0 and f ( ) = 0 Since f is decreasing on a neighborhood of, f ( ) < 0. Therefore, g ( ) < 0. : { : answer : justification (d) f ( ) f () g ( ) g ( ) e f ( ) e f () = = e : difference quotient : { : answer 009 The College Board. All rights reserved. Visit the College Board on the Web:

3 008 SCORING GUIDELINES (Form B) Question Let g be a continuous function with g ( ) =. The graph of the piecewise-linear function g, the derivative of g, is shown above for x 7. (a) Find the x-coordinate of all points of inflection of the graph of y = g for < x < 7. Justify your answer. (b) Find the absolute maximum value of g on the interval x 7. Justify your answer. (c) Find the average rate of change of g( x ) on the interval x 7. (d) Find the average rate of change of g on the interval x 7. Does the Mean Value Theorem applied on the interval x 7 guarantee a value of c, for < c < 7, such that g ( c) is equal to this average rate of change? Why or why not? (a) g changes from increasing to decreasing at x = ; g changes from decreasing to increasing at x = 4. : { : x -values : justification Points of inflection for the graph of y = g occur at x = and x = 4. (b) The only sign change of g from positive to negative in the interval is at x =. g = + = = g( ) = 7 g( 7) = + g dx = + ( 4) + = ( ) g dx ( ) : identifies x = as a candidate : considers endpoints : maximum value and justification The maximum value of g for x 7 is. (c) g( 7) g( ) = = 7 ( ) 0 : difference quotient : { : answer (d) g ( 7) g ( ) ( 4) = = 7 ( ) 0 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 < x < 7. : average value of g : : answer No with reason 008 The College Board. All rights reserved. Visit the College Board on the Web:

4 007 SCORING GUIDELINES (Form B) Let f be a function defined on the closed interval x with f ( ) =. The graph of f, the derivative of f, consists of two semicircles and two line segments, as shown above. (a) For < x <, find all values x at which f has a relative maximum. Justify your answer. (b) For < x <, find all values x at which the graph of f has a point of inflection. Justify your answer. (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 x. Explain your reasoning. (a) f = 0 at x =,, 4 f changes from positive to negative at and 4. Thus, f has a relative maximum at x = and at x = 4. : { : x -values : justification (b) f changes from increasing to decreasing, or vice versa, at x = 4,, and. Thus, the graph of f has points of inflection when x = 4,, and. : { : x -values : justification (c) The graph of f is concave up with positive slope where f is increasing and positive: < x < 4 and < x <. : { : intervals : explanation (d) Candidates for the absolute minimum are where f changes from negative to positive (at x = ) and at the endpoints ( x =, ). π f( ) = + f dx = + π > f ( ) = f( ) = + f dx = + > The absolute minimum value of f on [, ] is f ( ) =. : identifies x = as a candidate : considers endpoints : value and explanation 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

5 006 SCORING GUIDELINES (Form B) 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, f () t = t + t + for 4 0 t 4 and f is piecewise linear for 4 t 4. (a) Find f ( ). Indicate units of measure. (b) For the time interval 0 t 4, 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 t 8 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 calories per minute is burned during the time interval 6 t 8. (a) f ( ) = = 0 4 calories/min/min : f ( ) and units (b) f is increasing on [ 0, 4 ] and on [, 6 ]. 9 On (, 6 ), f () t = = since f has 6 constant slope on this interval. On ( 0, 4 ), f () t = t + t and 4 f () t = t + = 0 when t =. This is where f has a maximum on [ 0, 4 ] since f > 0 on ( 0, ) and f < 0 on (, 4 ). On [ 0, 4 ], f is increasing at its greatest rate when t = because f ( ) = >. 8 (c) f() t dt = 69 ( ) + ( 4)( 9+ ) + ( ) 6 = calories 8 (d) We want ( () ). f t + c dt = 6 This means + c = (). So, c = 4. OR Currently, the average is = calories/min. Adding c to f () t will shift the average by c. So c = 4 to get an average of calories/min. : f on ( 0, 4) : shows f has a max at t = on ( 0, 4) 4 : : shows for < t < 6, f () t < f ( ) : answer : { : method : answer : { : setup : value of c 006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents).

6 006 SCORING GUIDELINES (Form B) Question Let f be the function defined for x 0 with f ( 0) = and f, the ( x 4) first derivative of f, given by f = e sin ( x ). The graph of y = f is shown above. (a) Use the graph of f to determine whether the graph of f is concave up, concave down, or neither on the interval.7 < x <.9. Explain your reasoning. (b) On the interval 0 x, find the value of x at which f has an absolute maximum. Justify your answer. (c) Write an equation for the line tangent to the graph of f at x =. (a) On the interval.7 < x <.9, f is decreasing and thus f is concave down on this interval. : { : answer : reason (b) f = 0 when x = 0, π, π, π, K On [0, ] f changes from positive to negative only at π. The absolute maximum must occur at x = π or at an endpoint. f ( 0) = ( π ) ( ) ( ) f = f 0 + f x dx = f( ) = f( 0) + f dx = π : identifies π and as candidates - or - indicates that the graph of f increases, decreases, then increases : justifies f( π ) > f( ) : answer This shows that f has an absolute maximum at x = π. (c) f( ) = f( 0) + f dx =.64 f ( ) = e 0. sin( 4) = y.6 = ( 0.49)( x ) 0 : f ( ) expression : integral 4 : : including f ( 0 ) term : f ( ) : equation 006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents).

7 004 SCORING GUIDELINES (Form B) The figure above shows the graph of f, the derivative of the function f, on the closed interval x. The graph of f has horizontal tangent lines at x = and x =. The function f is twice differentiable with f ( ) = 6. (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 x? At what value of x does f attain its absolute maximum value on the closed interval x? Show the analysis that leads to your answers. (c) Let g be the function defined by g = xf. Find an equation for the line tangent to the graph of g at x =. (a) x = and x = because the graph of f changes from increasing to decreasing at x =, and changes from decreasing to increasing at x =. : x =, x = : : reason (b) The function f decreases from x = to x = 4, then increases from x = 4 to x =. Therefore, the absolute minimum value for f is at x = 4. The absolute maximum value must occur at x = or at x =. f( ) f( ) = f ( t) dt < 0 Since f( ) < f( ), the absolute maximum value occurs at x =. 4 : : indicates f decreases then increases : eliminates x = for maximum : absolute minimum at x = 4 : absolute maximum at x = (c) g = f + xf g ( ) = f( ) + f ( ) = 6 + ( ) = 4 g( ) = f( ) = : g : tangent line Tangent line is y = 4( x ) + Copyright 004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents).

AP CALCULUS AB 2004 SCORING GUIDELINES (Form B)

AP CALCULUS AB 2004 SCORING GUIDELINES (Form B) AP CALCULUS AB 004 SCORING GUIDELINES (Form B) Question 4 The figure above shows the graph of f, the derivative of the function f, on the closed interval 1 x 5. The graph of f has horizontal tangent lines