To approximate the cube root of 65 using the differential form, we start by defining the function f(x)=x3f(x) = \\sqrt[3]{x}f(x)=3x\u200b, which is the cube root of xxx.<\/p>\n\n\n\n
Step 1: Choose a point of approximation<\/h3>\n\n\n\n
Since we are approximating 653\\sqrt[3]{65}365\u200b, we choose a point close to 65 where the cube root is easy to calculate. A natural choice is x=64x = 64x=64, since 643=4\\sqrt[3]{64} = 4364\u200b=4.<\/p>\n\n\n\n
Step 2: Compute the derivative of f(x)f(x)f(x)<\/h3>\n\n\n\n
We need the derivative of f(x)=x1\/3f(x) = x^{1\/3}f(x)=x1\/3. Using the power rule, the derivative is:f\u2032(x)=13x\u22122\/3f'(x) = \\frac{1}{3} x^{-2\/3}f\u2032(x)=31\u200bx\u22122\/3<\/p>\n\n\n\n
Step 3: Apply the differential formula<\/h3>\n\n\n\n
The differential approximation formula is:f(x+\u0394x)\u2248f(x)+f\u2032(x)\u0394xf(x + \\Delta x) \\approx f(x) + f'(x) \\Delta xf(x+\u0394x)\u2248f(x)+f\u2032(x)\u0394x<\/p>\n\n\n\n
Where:<\/p>\n\n\n\n
- \n
- f(x)=x3f(x) = \\sqrt[3]{x}f(x)=3x\u200b is the original function.<\/li>\n\n\n\n
- f\u2032(x)=13x\u22122\/3f'(x) = \\frac{1}{3} x^{-2\/3}f\u2032(x)=31\u200bx\u22122\/3 is the derivative.<\/li>\n\n\n\n
- \u0394x=65\u221264=1\\Delta x = 65 – 64 = 1\u0394x=65\u221264=1 is the change in xxx.<\/li>\n<\/ul>\n\n\n\n
Step 4: Calculate the approximation<\/h3>\n\n\n\n
At x=64x = 64x=64, we know that f(64)=4f(64) = 4f(64)=4. Now, we evaluate f\u2032(64)f'(64)f\u2032(64):f\u2032(64)=13\u00d764\u22122\/3f'(64) = \\frac{1}{3} \\times 64^{-2\/3}f\u2032(64)=31\u200b\u00d764\u22122\/3<\/p>\n\n\n\n
Since 641\/3=464^{1\/3} = 4641\/3=4, 64\u22122\/3=142=11664^{-2\/3} = \\frac{1}{4^2} = \\frac{1}{16}64\u22122\/3=421\u200b=161\u200b. Thus:f\u2032(64)=13\u00d7116=148f'(64) = \\frac{1}{3} \\times \\frac{1}{16} = \\frac{1}{48}f\u2032(64)=31\u200b\u00d7161\u200b=481\u200b<\/p>\n\n\n\n
Now, we can approximate f(65)f(65)f(65) as:f(65)\u2248f(64)+f\u2032(64)\u00d7\u0394x=4+148\u00d71=4+148f(65) \\approx f(64) + f'(64) \\times \\Delta x = 4 + \\frac{1}{48} \\times 1 = 4 + \\frac{1}{48}f(65)\u2248f(64)+f\u2032(64)\u00d7\u0394x=4+481\u200b\u00d71=4+481\u200b<\/p>\n\n\n\n
So:f(65)\u22484+0.0208=4.0208f(65) \\approx 4 + 0.0208 = 4.0208f(65)\u22484+0.0208=4.0208<\/p>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n
The approximation for 653\\sqrt[3]{65}365\u200b using the differential form is 4.02084.02084.0208.<\/p>\n\n\n\n
Explanation:<\/h3>\n\n\n\n
In this method, the differential approximation is based on linearizing the function at a point where the value is easy to calculate. By using the derivative to estimate how much the function changes as xxx increases from 64 to 65, we can get a very close approximation for 653\\sqrt[3]{65}365\u200b, which in this case is approximately 4.0208.<\/p>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"‘Use differential form to approximate %65 (cube root)-_’ The Correct Answer and Explanation is: To approximate the cube root of 65 using the differential form, we start by defining the function f(x)=x3f(x) = \\sqrt[3]{x}f(x)=3x\u200b, which is the cube root of xxx. Step 1: Choose a point of approximation Since we are approximating 653\\sqrt[3]{65}365\u200b, we choose […]<\/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-242694","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/242694","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=242694"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/242694\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=242694"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=242694"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=242694"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}