Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.1 Maximum and Minimum Values - 3.1 Execises - Page 212: 50


Absolute maximum: $f(3)=125$ Absolute minimum: $f(0)=-64$.

Work Step by Step

Using closed interval method, as $f$ in continuous over the given range, let's first calculate the value of $f$ at critical numbers followed by values of $f$ at endpoints of the given range. For critical numbers, $f'(t)=0$ or $f'(t)$ should not exist. Let's differentiate $f(t)=(t^2-4)^3$ with respect to $t$ keeping $[-2,3]$ as interval. $$\frac{df}{dt}=3(t^2-4)^2(2t)=6t(t^2-4)^2$$ $\frac{df}{dt}$ exists on the given interval. Thus, $\frac{df}{dt}=0$, $$6t(t^2-4)^2=0$$ Thus, $t=0$ and $(t^2-4)^2=0\implies t=-2,2$. Thus, $t=-2,0,2$ are critical numbers. Thus, $f(0)=-64, f(-2)=f(2)=0$. At endpoints, $f(-2)=0$ & $f(3)=125$. Thus, absolute maximum $=f(3)=125$ and absolute minimum $=f(0)=-64$.
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