Use the Alternate form of the Limit Definition of Derivative to evaluate the derivative at the specified x-value.<\/p>\n\n\n\n
a. f(x) = 4 – x ^ 2 at x = 3 f^ prime ( epsilon)+lim x+c (f(x) – f(epsilon))\/(x – c) \\ bm; x + 3 (- (x ^ 2 – a))\/(x – 3) f’ * (3) = lim x -> 3 ((9x ^ 4) – f\/2 * (3))\/(x – 3) matrix 4\\ 3 matrix (- (x + 3) * (x – 3))\/(x – 3) matrix nm = x + 3 lim x+3 (9 – x ^ 2 – (4 – 3 ^ 3))\/(x – 3) lim\\ x + 3 – (x + 3) (9 – x ^ 2 – 4 + 9)\/(x – 3) -(343) (9 – x ^ 2 – 4 + 9)\/(x – 3) (- x ^ x + 9)\/(x – 1) c. f(x) = |x + 5| x = – 5 (Hint: You will need to make use of one-sided limits.) f’ * c = lim x -> c (f(x) – f(c))\/(x – c) lim x -5 ^ – |x+5| x+5 lim x -5^ + (1x + 51)\/(x + 5) f^ prime (-5)=lim x -5 [|x + 5|] – f(- 5) x-(*5) matrix |-5^ * +5| -5^ * +5 |-5^ * -5| -5^ * +5 \\ |- 5 + 5|\/8 |-5^ * -5| 9^ * aligned lim v-5 [1×451]-0 x5 lim x -> – 5 |x + 5|\/(x + 5) SPUT THE LIMIT -1 and 1\u2026 f is not differentiable at y -5<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s break down the problem step by step and apply the limit definition of the derivative to each case. The derivative of a function ( f(x) ) at a specific point ( x = c ) is given by the limit definition of the derivative:<\/p>\n\n\n\n [ The function is ( f(x) = 4 – x^2 ), and we are asked to find the derivative at ( x = 3 ).<\/p>\n\n\n\n First, calculate ( f(3) ): Now, applying the limit definition: Factor the numerator: Thus, the expression becomes: Notice that we can cancel out the ( (x – 3) ) term (since ( x \\neq 3 )): Now substitute ( x = 3 ): So, the derivative of ( f(x) ) at ( x = 3 ) is ( \\boxed{-6} ).<\/p>\n\n\n\n We are asked to find the derivative of ( f(x) = |x + 5| ) at ( x = -5 ). Since ( f(x) = |x + 5| ), the function is piecewise and behaves differently depending on whether ( x + 5 ) is positive or negative.<\/p>\n\n\n\n Case 1: ( x > -5 )<\/strong><\/p>\n\n\n\n For ( x > -5 ), we have ( |x + 5| = x + 5 ). The derivative in this region is: Case 2: ( x < -5 )<\/strong><\/p>\n\n\n\n For ( x < -5 ), we have ( |x + 5| = -(x + 5) ). The derivative in this region is: At ( x = -5 ), the function is not differentiable.<\/strong><\/p>\n\n\n\n To show this, we use the one-sided limits:<\/p>\n\n\n\n Since the left-hand and right-hand limits are not equal, ( f'(x) ) does not exist at ( x = -5 ).<\/p>\n\n\n\n Thus, the derivative does not exist at ( x = -5 ), and we conclude that ( f(x) ) is not differentiable at ( x = -5 )<\/strong>.<\/p>\n\n\n\n This demonstrates the importance of considering one-sided limits when dealing with piecewise functions or absolute value functions.<\/p>\n\n\n\n <\/p>\n","protected":false},"excerpt":{"rendered":" Use the Alternate form of the Limit Definition of Derivative to evaluate the derivative at the specified x-value. a. f(x) = 4 – x ^ 2 at x = 3 f^ prime ( epsilon)+lim x+c (f(x) – f(epsilon))\/(x – c) \\ bm; x + 3 (- (x ^ 2 – a))\/(x – 3) 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-183606","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183606","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=183606"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183606\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183606"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183606"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183606"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
f'(c) = \\lim_{x \\to c} \\frac{f(x) – f(c)}{x – c}
]<\/p>\n\n\n\nProblem (a) ( f(x) = 4 – x^2 ) at ( x = 3 )<\/h3>\n\n\n\n
[
f(3) = 4 – 3^2 = 4 – 9 = -5
]<\/p>\n\n\n\n
[
f'(3) = \\lim_{x \\to 3} \\frac{f(x) – f(3)}{x – 3} = \\lim_{x \\to 3} \\frac{(4 – x^2) – (-5)}{x – 3}
]
[
f'(3) = \\lim_{x \\to 3} \\frac{4 – x^2 + 5}{x – 3} = \\lim_{x \\to 3} \\frac{9 – x^2}{x – 3}
]<\/p>\n\n\n\n
[
9 – x^2 = (3 – x)(3 + x)
]<\/p>\n\n\n\n
[
f'(3) = \\lim_{x \\to 3} \\frac{(3 – x)(3 + x)}{x – 3}
]<\/p>\n\n\n\n
[
f'(3) = \\lim_{x \\to 3} -(3 + x)
]<\/p>\n\n\n\n
[
f'(3) = -(3 + 3) = -6
]<\/p>\n\n\n\n
\n\n\n\nProblem (b) ( f(x) = |x + 5| ) at ( x = -5 )<\/h3>\n\n\n\n
[
f'(x) = \\frac{d}{dx}(x + 5) = 1
]<\/p>\n\n\n\n
[
f'(x) = \\frac{d}{dx}(-(x + 5)) = -1
]<\/p>\n\n\n\n\n
[
\\lim_{x \\to -5^-} \\frac{|x + 5| – |0|}{x + 5} = \\lim_{x \\to -5^-} \\frac{-(x + 5)}{x + 5} = -1
]<\/li>\n\n\n\n
[
\\lim_{x \\to -5^+} \\frac{|x + 5| – |0|}{x + 5} = \\lim_{x \\to -5^+} \\frac{x + 5}{x + 5} = 1
]<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nConclusion<\/h3>\n\n\n\n
\n