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: 45


Absolute maximum: $f(2)=16$ Absolute minimum: $f(5) = 7$

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, $f'(x)=0$ or $f'(x)$ should not exist. Let's differentiate $f(x)=12+4x-x^2$ with respect to $x$ keeping $[0,5]$ as interval. $$\frac{df}{dx}=4-2x$$ $\frac{df}{dx}$ exists on the given interval. Thus, $\frac{df}{dx}=0$, $$4-2x=0\implies x=2.$$ Thus, $f(2)=12+4(2)-2^2=16$. At endpoints, $f(0)=12$ & $f(5)=7$. Thus, absolute maximum $=f(2)=16$ and absolute minimum $=f(5)=7$.
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