To solve this problem, we need to linearize the cube root function at x=64x = 64x=64. The general approach is to use the tangent line to the function f(x)=x3f(x) = \\sqrt[3]{x}f(x)=3x\u200b at x=64x = 64x=64, which will serve as the linear approximation, L(x)L(x)L(x).<\/p>\n\n\n\n
Step 1: Find the cube root function and its derivative<\/h3>\n\n\n\n
We are given the cube root function:f(x)=x3f(x) = \\sqrt[3]{x}f(x)=3x\u200b<\/p>\n\n\n\n
The derivative of f(x)f(x)f(x), using the power rule, is:f\u2032(x)=ddx(x1\/3)=13x\u22122\/3=13x23f'(x) = \\frac{d}{dx} \\left( x^{1\/3} \\right) = \\frac{1}{3} x^{-2\/3} = \\frac{1}{3 \\sqrt[3]{x^2}}f\u2032(x)=dxd\u200b(x1\/3)=31\u200bx\u22122\/3=33×2\u200b1\u200b<\/p>\n\n\n\n
Step 2: Linearize f(x)f(x)f(x) at x=64x = 64x=64<\/h3>\n\n\n\n
We need the equation of the tangent line to f(x)f(x)f(x) at x=64x = 64x=64. This tangent line is given by the linear approximation formula:L(x)=f(a)+f\u2032(a)(x\u2212a)L(x) = f(a) + f'(a)(x – a)L(x)=f(a)+f\u2032(a)(x\u2212a)<\/p>\n\n\n\n
where a=64a = 64a=64.<\/p>\n\n\n\n
- \n
- Find f(64)f(64)f(64):<\/strong> f(64)=643=4f(64) = \\sqrt[3]{64} = 4f(64)=364\u200b=4<\/li>\n\n\n\n
- Find f\u2032(64)f'(64)f\u2032(64):<\/strong> f\u2032(64)=136423=13\u22c516=148f'(64) = \\frac{1}{3 \\sqrt[3]{64^2}} = \\frac{1}{3 \\cdot 16} = \\frac{1}{48}f\u2032(64)=33642\u200b1\u200b=3\u22c5161\u200b=481\u200b<\/li>\n<\/ol>\n\n\n\n
Now the linear approximation L(x)L(x)L(x) is:L(x)=4+148(x\u221264)L(x) = 4 + \\frac{1}{48} (x – 64)L(x)=4+481\u200b(x\u221264)<\/p>\n\n\n\n
Step 3: Approximate the cube root of 65<\/h3>\n\n\n\n
We now use this linear approximation to approximate the cube root of 65. Substituting x=65x = 65x=65 into the linearization formula:L(65)=4+148(65\u221264)=4+148L(65) = 4 + \\frac{1}{48} (65 – 64) = 4 + \\frac{1}{48}L(65)=4+481\u200b(65\u221264)=4+481\u200b<\/p>\n\n\n\n
Thus, the cube root of 65 is approximately:L(65)=4+148L(65) = 4 + \\frac{1}{48}L(65)=4+481\u200b<\/p>\n\n\n\n
Step 4: Find how much the answer is greater than 4<\/h3>\n\n\n\n
The difference between the linear approximation and 4 is:L(65)\u22124=148L(65) – 4 = \\frac{1}{48}L(65)\u22124=481\u200b<\/p>\n\n\n\n
Thus, the answer is 148\\frac{1}{48}481\u200b<\/strong>, which means the linear approximation is greater than 4 by 148\\frac{1}{48}481\u200b. The correct answer is:148\\boxed{\\frac{1}{48}}481\u200b\u200b<\/p>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"‘Let L() be the linearization of the cube root function at X-64. When using L(x) to approximate the cube root of 65, how much is the answer greater than 4? 1\/4 1\/48 1\/65 1\/64 21\/3’ The Correct Answer and Explanation is: To solve this problem, we need to linearize the cube root function at x=64x […]<\/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-242678","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/242678","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=242678"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/242678\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=242678"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=242678"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=242678"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Find f\u2032(64)f'(64)f\u2032(64):<\/strong> f\u2032(64)=136423=13\u22c516=148f'(64) = \\frac{1}{3 \\sqrt[3]{64^2}} = \\frac{1}{3 \\cdot 16} = \\frac{1}{48}f\u2032(64)=33642\u200b1\u200b=3\u22c5161\u200b=481\u200b<\/li>\n<\/ol>\n\n\n\n