![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/01\/image-127.png\")
<\/figure>\n\n\n\nThe Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve this problem, we need to identify the local and absolute extreme values of the function ( g(t) = t^3 – 112t^2 + 28t ) within the given domain ( 0 \\leq t \\leq ) (it’s likely that the upper bound for ( t ) is missing).<\/p>\n\n\n\n We begin by finding the first derivative of the function to locate the critical points, which occur where the first derivative equals zero or is undefined.<\/p>\n\n\n\n [ Set the first derivative equal to zero to find the critical points:<\/p>\n\n\n\n [ We solve this quadratic equation using the quadratic formula:<\/p>\n\n\n\n [ This gives us two critical points: Now, find the second derivative to determine whether the critical points are maxima or minima.<\/p>\n\n\n\n [ To classify the critical points, substitute ( t_1 ) and ( t_2 ) into the second derivative:<\/p>\n\n\n\n Next, evaluate the function at the critical points and the endpoints ( t = 0 ) and ( t = 74.53 ) (if this is the domain boundary):<\/p>\n\n\n\n Identify the function’s extreme values in the given domain, and give the coordinates where they are located. Identify all local and all absolute maxima and minima. All work steps on scrap, answers on test. g(t)=t33-112t2+28t,0=t<> local minimum: absolute minimum: local maximum: absolute maximim: For the toolbar, press ALT+F10 (PC) ?or ALT+FN+F10 (Mac). The Correct Answer […]<\/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-181871","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181871","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=181871"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181871\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=181871"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=181871"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=181871"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Find the first derivative<\/h3>\n\n\n\n
g'(t) = \\frac{d}{dt}(t^3 – 112t^2 + 28t) = 3t^2 – 224t + 28
]<\/p>\n\n\n\nStep 2: Find the critical points<\/h3>\n\n\n\n
3t^2 – 224t + 28 = 0
]<\/p>\n\n\n\n
t = \\frac{-(-224) \\pm \\sqrt{(-224)^2 – 4(3)(28)}}{2(3)}
]
[
t = \\frac{224 \\pm \\sqrt{50176 – 336}}{6}
]
[
t = \\frac{224 \\pm \\sqrt{49840}}{6}
]
[
t = \\frac{224 \\pm 223.18}{6}
]<\/p>\n\n\n\n
[
t_1 = \\frac{224 + 223.18}{6} \\approx 74.53 \\quad \\text{and} \\quad t_2 = \\frac{224 – 223.18}{6} \\approx 0.14
]<\/p>\n\n\n\nStep 3: Determine the second derivative<\/h3>\n\n\n\n
g”(t) = \\frac{d}{dt}(3t^2 – 224t + 28) = 6t – 224
]<\/p>\n\n\n\nStep 4: Test the critical points<\/h3>\n\n\n\n
\n
[
g”(74.53) = 6(74.53) – 224 \\approx 447.18 – 224 = 223.18 \\quad (\\text{positive, local minimum})
]<\/li>\n\n\n\n
[
g”(0.14) = 6(0.14) – 224 \\approx 0.84 – 224 = -223.16 \\quad (\\text{negative, local maximum})
]<\/li>\n<\/ul>\n\n\n\nStep 5: Evaluate function at critical points and endpoints<\/h3>\n\n\n\n
\n
[
g(74.53) \\approx 411926.55 – 612497.97 + 2087.84 \\approx -199483.58
]<\/li>\n\n\n\n
[
g(0.14) \\approx 0.002744 – 2.2208 + 3.92 \\approx 1.701
]<\/li>\n<\/ul>\n\n\n\nStep 6: Conclusion<\/h3>\n\n\n\n
\n