2014 AP” CALCULUS AB FREE-RESPONSE QUESTIONS<\/p>\n\n\n\n
x-2-2<x<-1-1-1<x<11<x<33 f(x) 12 Positive 8 Positive 2 Positive 7 f'(x)-5 Negative 0 Negative 0 Positive 1 2 g(x) -1 Negative 0 Positive 3 Positive 1 3 Positive Positive 0 Negative-2 2 g'(x) 2 5 The twice-differentiable functions f and gare defined for all real numbers x. Values of f, f’, g, and g’ for various values of x are given in the table above. (a) Find the x-coordinate of each relative minimum of f on the interval [-2,3]. Justify your answers. (b) Explain why there must be a value c, for-1 <c< 1, such that f” (c) = 0. (c) The function h is defined by h(x) = ln(f(x)). Find h’ (3). Show the computations that lead to your answer. (d) Evaluate [f'(g(x))g'(x)dx<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n A relative minimum occurs where ( f'(x) ) changes from negative to positive, indicating a change in the direction of ( f(x) ).<\/p>\n\n\n\n Since ( f'(x) ) changes from negative to positive at ( x = 0 ), ( x = 0 ) is the location of a relative minimum.<\/p>\n\n\n\n Answer<\/strong>: The relative minimum of ( f(x) ) occurs at ( x = 0 ).<\/p>\n\n\n\n The function ( f(x) ) is twice differentiable (and thus continuous). Between (-1 < x < 1), ( f'(x) ) changes from ( -5 ) at ( x = -1 ) to ( 1 ) at ( x = 1 ). By the Mean Value Theorem, ( f'(x) ) must have an intermediate slope value of ( 0 ) within this interval. Since ( f”(x) ) represents the rate of change of ( f'(x) ), ( f”(c) = 0 ) for some ( c ) in ((-1, 1)).<\/p>\n\n\n\n By the chain rule: Answer<\/strong>: ( h'(3) = \\frac{2}{7} ).<\/p>\n\n\n\n Let ( u = g(x) ). Then ( du = g'(x) dx ), and the integral becomes: Answer<\/strong>: ( f(g(x)) + C ).<\/p>\n\n\n\n In (a), relative extrema occur where ( f'(x) ) changes sign. Observing the sign change at ( x = 0 ), it indicates a relative minimum. For (b), the continuity and differentiability of ( f(x) ) ensure the conditions of the Mean Value Theorem are met. Since ( f'(x) ) transitions from negative to positive, ( f”(x) = 0 ) must occur within the interval. For (c), the logarithmic derivative rule, ( h'(x) = \\frac{f'(x)}{f(x)} ), simplifies the calculation at ( x = 3 ) using the provided table values. Finally, in (d), the substitution ( u = g(x) ) transforms the integral, leveraging the fundamental theorem of calculus to yield ( f(u) + C ).<\/p>\n","protected":false},"excerpt":{"rendered":" 2014 AP” CALCULUS AB FREE-RESPONSE QUESTIONS x-2-2<x<-1-1-1<x<11<x<33 f(x) 12 Positive 8 Positive 2 Positive 7 f'(x)-5 Negative 0 Negative 0 Positive 1 2 g(x) -1 Negative 0 Positive 3 Positive 1 3 Positive Positive 0 Negative-2 2 g'(x) 2 5 The twice-differentiable functions f and gare defined for all real numbers x. Values of f, […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-185286","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185286","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=185286"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185286\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185286"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185286"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185286"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Problem Breakdown and Solution:<\/h3>\n\n\n\n
(a) Finding the x-coordinate of each relative minimum of ( f(x) ) on the interval ([-2, 3]):<\/h4>\n\n\n\n
\n
(b) Explaining why there must be a value ( c ) such that ( f”(c) = 0 ) for (-1 < c < 1):<\/h4>\n\n\n\n
(c) Finding ( h'(3) ), where ( h(x) = \\ln(f(x)) ):<\/h4>\n\n\n\n
[
h'(x) = \\frac{f'(x)}{f(x)}.
]
At ( x = 3 ), the table gives ( f'(3) = 2 ) and ( f(3) = 7 ). Therefore:
[
h'(3) = \\frac{f'(3)}{f(3)} = \\frac{2}{7}.
]<\/p>\n\n\n\n(d) Evaluating (\\int f'(g(x)) g'(x) \\, dx):<\/h4>\n\n\n\n
[
\\int f'(g(x)) g'(x) \\, dx = \\int f'(u) \\, du.
]
The integral of ( f'(u) ) is ( f(u) + C ), where ( u = g(x) ). Thus:
[
\\int f'(g(x)) g'(x) \\, dx = f(g(x)) + C.
]<\/p>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/h3>\n\n\n\n