AP Eam Practice Questions for Chapter AP Eam Practice Questions for Chapter f 4 + 6 7 9 f + 7 0 + 6 0 ( + )( ) 0,. The critical numbers of f( ) are and.. Evaluate each point. A: d d C: d d B: D: d d d d. st () t s () t t + t + 9t+ 5 + 6t + 9 s t 6t + 6 () () t 0 s 6t + 6 0 6t 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. 4. y + y y 44 y ± y > y 44 Because 0, use 44. Let A be the area to be maimized. A y 44, 0 < < ( 44 ) da 44 ( 44 ) ( ) 44 d + 44 da d 0 44 44 44 ± 7 ± 6 Because, use 6. y 44 6 7 6 A 6 6 6 square units. The maimum area is 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
AP Eam Practice Questions for Chapter 5. Evaluate each statement. A: The point ( 4, ) 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 (, 4 ), but there may be a point on (, 4) at which g ( ) 0. The statement may not be true. 6. Evaluate each statement. A: f ( 0) may or may not be undefined based on the function f. The statement may or may not be true. lim f may or may not eist based on the 0 function f. The statement may or may not be true. B: C: Because y 0 lim f 0. is a horizontal asymptote, The statement must be true. D: Even though y 0 is a horizontal asymptote, there may be at least one value of for which f( ) 0. The statement may or may not be true. 7. Evaluate each statement. lim f and f 4 8, f is not continuous at 4. I. Because II. f( ) 4 The statement is true. 4 + + 4 lim lim lim 4 8 8 The statement is false. + 4 ( + 8)( 4) + 8 f, 4 8 4 + + III. f has a removable discontinuity at 4, not a vertical asymptote. The statement is false. Because I is the only statement that is true, the answer is A. 8. y f() c + f ()( c c) y f() + f ()( ) y 8 + ( ) y 58 f (.9) (.9) 58 5.8 f + 4 + cos 9. f + 8 + + sin f 6 + 8 + cos f ( ) 6 + 8 + cos 0 when.67, which is the only point of inflection of the graph of f. 0. lim lim lim + 4 4 4 0 + + 6 6 4 6 The answer is D. 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
AP Eam Practice Questions for Chapter. (a) f( ) sin + cos f Using a graphing utility, f ( ) cos 0 when ± 0.708. So, the relative etrema of f occur at ± 0.708. Notes: 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. Round each answer to at least three decimal places to receive credit on the eam. (b) f sin + 6 + 6 f cos 8 Tangent line: y 6 8 y + 8 6 6 y 8 8 (c) Using the tangent line approimation, f (.5) (.5).676. 8 8 The actual value of (.5) f.5 sin.5 +.690. So, the tangent line approimation is an f.5. underestimate of Notes: 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 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. 08 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. (a) Because f ( ) when < 4, f is increasing on the interval (, 4 ). (b) Yes. Because f is continuous and f ( ) changes from positive to negative at 4, f has a relative maimum at 4. pts: answer with justification Note: In such an eplanation, use in your justification (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 4, f ( ) on (, 4 ), and f ( ) 0 on ( 4, ), > the point of inflection is at 4. pts: answer with justification at Note: Be sure to use in your justification (eplain using a derivative) rather than reasoning with the concavity of f. (d) No, f ( ) is not differentiable on (, 5 ). pt: answer with justification (e) Answers will vary. Sample answer: y 6 5 4 (, 4) y 0 4 5 6 7 8 9 Note: 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. 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
AP Eam Practice Questions for Chapter 5. (a) Because f ( ) when 0 < <, f is increasing on the interval ( 0, ). pts: answer with justification Note: In your justification, 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 0.8 0 and. 0, the graph of f has points of inflection at 0.8.. The graph of f is decreasing when and < < 0.8 and >., so the graph of f is.,. concave downward on (, 0.8) and Note: In your justification, be sure to eplicitly identify each function by name. (c) By the Mean Value Theorem, there eists a number c in ( 0.5, 0) such that f( 0.5) f( 0) f () c. 0.5 0 Because f () c for 0.5 < c, f( 0.5) f( 0) 0.5 0. pts: answer with justification 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. 08 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 AP Eam Practice Questions for Chapter 4. (a) f ( ) 4 4 f ( ) 4 f Use a table to test There are no points at which 0. the critical number 0, where f does not eist. Interval < 0 < < Test value Sign of f ( ) is So, f is decreasing on (, 0 ) and ( 0, ) because f ( ) negative on these intervals. f f () 5 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. < (b) f is concave downward when f ( ) 0. f ( ) 4 f ( ) f ( ) when. So, on (, 0 ). f is concave downward (c) Because f ( ) 0 and f ( ) does not eist when 0, the graph of f does not have any points of inflection. pt: answer with justification Note: In these justifications, 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. 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
AP Eam Practice Questions for Chapter 7 5. (a) Use f f f ( ) on [, 0 ]. 0 and 0 to find an equation of 0 m 0 f ( ) 0 ( ) f ( ) f. So, at,. Because m f f (b) On the interval f 5, 0, 0. So, f has a critical number at. Because f ( ) on ( 5, ) and f ( ) < ( ) f has a relative maimum at the point where. 0 on, 0, pts: answer with justification changes from positive to negative at (c) The points of inflection occur at 4, 0, and because f ( 4) 0, f ( 0 ) and f ( ) and f ( ) are undefined, changes from either increasing to decreasing or decreasing to increasing at these -values [see part (c)]. pts: answers with justification or where is undefined and that changes from increasing to decreasing or from decreasing to increasing at these (d) g f + sin g f + sin cos g f + sin cos 4 4 4 4 f + f 4 4 From the graph, f 4 is negative. So, g 4 is negative, which means that g is decreasing at. 4 Note: In these justifications, be sure to eplicitly identify each function by name. For eample, referring to as it, the function, or the graph may not receive credit on the eam. 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.