AP Exam Practice Questions for Chapter 3

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1 AP Eam Practice Questions for Chapter AP Eam Practice Questions for Chapter f f ( + )( ) 0,. The critical numbers of f are and. So, the answer is B.. Evaluate each statement. I: Because the function is strictly monotonic and increasing, f > 0 on the entire number line. The statement is true. II: Because f is concave downward when <, < 0 on the interval (, ). f The statement is true. III: Because f is concave upward when >, > 0 on the interval (, ). f The statement is true. So, the answer is D.. st () t s () t t + t + 9t t + 9 s t 6t + 6 () () t 0 s 6t t 6 t So, t is a point of inflection of s(). t Use s () t to find the velocity at t. s () () + 6 () + 9 The maimum velocity is feet per second. So, the answer is B.. Evaluate each statement. A: The point (, ) appears to be a relative minimum, but there may be another number c on [, 6] for which g () c 0. The statement may not be true. B: The point ( 6, 7) appears to be a relative maimum, but there may be another number c on [, 6] for which g () c 0. The statement may not be true. C: Because g is continuous and differentiable on [, 6] and g g( 6 ), in (, 6) such that g () c 0. then there is at least one number c By Rolle s Theorem, this statement must be true. D: The graph of g appears to be decreasing on (, ), but there may be a point on (, ) at which g 0. The statement may not be true. So, the answer is C. 5. f 0 f < 0because f is decreasing at. f > 0 because the graph of f is concave upward at. Therefore, f < f < f f > f( ). So, the answer is D. 6. y arctan Let u du d. du dy + u d + d + 6 So, the answer is A. 07 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2 AP Eam Practice Questions for Chapter 7. f ln( ) f f( b) f( a) f() 8 f() ln 5 ln ln 5 By the Mean Value Theorem, f () c b a 8 f f ( c) ln 5 ln 5 + ln So, the answer is A. 8. (a) Because f > 0 when <, f is increasing on the interval (, ). (b) Yes. Because f is continuous and f changes from positive to negative at, f has a relative maimum at. pts: answer with reason Reminder: In such an eplanation, use in your reason (eplain using a derivative). Merely reasoning with f (appealing to where f changes from increasing to decreasing) may not receive credit on the eam. (c) Because f is continuous, f, f < 0 on (, ), and f 0 on (, ), > the point of inflection is at. pts: answer with reason at Reminder: Be sure to use in your reason (eplain using a derivative) rather than reasoning with the concavity of f. (d) No, f is not differentiable on (, 5 ). pt: answer with reason (e) Answers will vary. Sample answer: y 6 5 (, ) y Reminder: In the eplanations throughout this question, be sure to eplicitly identify each function by name. For eample, referring to in part (a) as it or the function may not receive credit on the eam because there are three functions involved in the analysis of this question. 07 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3 AP Eam Practice Questions for Chapter 9. (a) f sin + cos f Using a graphing utility, f cos 0 when ± So, the relative etrema of f occur at ± Note: You would be epected to compute/work with here in justifying the critical points. Merely obtaining the critical points from your calculator would not receive full credit on the eam. Reminder: Round each answer to at least three decimal places to receive credit on the eam. (b) f sin f cos 8 Tangent line: y 6 8 y y 8 8 (c) Using the tangent line approimation, f (.5) (.5) The actual value of (.5) f.5 sin So, the tangent line approimation is an f.5. underestimate of Note: An alternate eplanation may be to identify that the tangent line at is below the graph of f. To see this, analyze the sign of to determine the concavity of f at Reminders: When using the tangent line to approimate be sure to write rather than Because this is an approimation, a point may be deducted if an equal sign is used. In general, equating two quantities that are not truly equal will result in a one point deduction on a free-response question. Be sure to round each answer to at least three decimal places to receive credit on the eam, and avoid premature rounding in intermediate steps. Be sure your calculator is in radian mode. 07 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4 AP Eam Practice Questions for Chapter 0. (a) f + cos f + ( sin ) sin 0 sin sin Check : f + cos + cos Check the endpoints of [ 0, ]: f ( 0) ( 0) + cos( 0) cos 0 f + cos + 0, the maimum value of f is +. Because f < f < f( ) (b) Because f is continuous on [ 0, ] and differentiable on ( 0, ), the conditions for the Mean Value Theorem are satisfied. f ( c) f f( 0) ( + ) 0 f f ( c) sin sin 0 sin 0 By the Mean Value Theorem,. 07 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5 AP Eam Practice Questions for Chapter 5. (a) Because f > 0 when 0 < <, f is increasing on the interval ( 0, ). pts: answer with reason Reminder: In your reason, be sure to eplicitly identify each function by name. Referring to as it, the function, or the graph may not receive credit on the eam. (b) Because f f and. 0, the graph of f has points of inflection at The graph of f is decreasing when and < < 0.8 and >., so the graph of f is concave downward on (, 0.8) and (., ). Reminder: In your reason, be sure to eplicitly identify each function by name. (c) By the Mean Value Theorem, f( 0.5) f( 0) f ( 0.5 ) Because f ( 0.5) < 0, the slope of f is negative at 0.5. pts: answer with reason Note: In addition to reasoning with the Mean Value Theorem, an alternate eplanation may involve justifying the sign of the numerator in the given difference quotient. Because on from the given graph, f is decreasing on this interval. So, and the numerator of this difference quotient must be positive. With a positive numerator and negative denominator, the difference quotient itself must be negative. 07 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6 6 AP Eam Practice Questions for Chapter. (a) f f f < f is decreasing when 0. 0 ± Use a table to test the critical numbers where Interval f does not eist. < < Test value Sign of f f < 0 ± and 0, < < 0 f 0 < 0 Graph of f Decreasing Decreasing Notes: For the justification in this particular eample, you could simply identify that is negative for all nonzero -values. A sign chart may not be necessary. If using a sign chart as part of the justification, the functions must be eplicitly labeled in your chart. Unlabeled sign charts may not receive credit on the eam. A sign chart alone is generally not sufficient for the eplanation. To receive full credit on the eam, be sure to eplain the information contained in the sign chart. Interval Test value Sign of f is So, f is decreasing on (, 0 ) and ( 0, ) because f negative on these intervals. 0 < < < < f 0 < 0 f () 5 < 0 Graph of f Decreasing Decreasing < (b) f is concave downward when f 0. f f f < 0 when < 0. So, on (, 0 ). f is concave downward (c) Because f 0, points of inflection. the graph of f does not have any pt: answer with reason Reminder: In these eplanations, be sure to eplicitly identify each function by name. Referring to or as it, the function, or the graph may not receive credit on the eam. 07 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7 AP Eam Practice Questions for Chapter 7. (a) Use f f f on [, 0 ]. 0 and 0 to find an equation of 0 m 0 ( ) y 0 y f. m at f, f. So, Because (b) On the interval f 5, 0, 0. So, the graph has a critical number at. Because the interval is open, the endpoints cannot be relative etrema. So, is a relative maimum because f > 0 on ( 5, ) and f < 0 on (, 0 ). (c) Possible points of inflection occur at, 0, and because f ( ) 0 and f ( 0 ) and f are undefined. When 5 < <, f is increasing. When < < 0, f is decreasing. When 0 < <, f is increasing. When < <, f is decreasing. So, f is concave downward on the intervals (, 0) and (, ) because f is decreasing. (d) The points of inflection occur at, 0, and because f ( ) 0, f ( 0 ) and f are undefined, and f changes from either increasing to decreasing or decreasing to increasing at these -values [see part (c)]. ( ) ( pt: finds f with justification finds/uses an equation of the given line segment) pts: pt: finds f ( ) with justification ( finds/uses the slope of the given line segment) pt: answer with reason identifies that f sign at ] changes identifies where < 0 ] pts: answers with reason [ f by eamining where f is decreasing identifies where 0 pts: answers with reason [ f or where f is undefined and that f changes from increasing to decreasing or from decreasing to increasing at these -values] (e) g f + sin g f + sin cos g f + sin cos f + f From the graph, f is negative. So, g is negative, which means that g is decreasing at. pt: computes g pts: pt: answer with reason identifies that g is negative Reminder: In these eplanations, be sure to eplicitly identify each function by name. For eample, referring f or f as it, the function, or the graph to may not receive credit on the eam. 07 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.