## Calculus 8th Edition

Published by Cengage

# Chapter 3 - Applications of Differentiation - 3.1 Maximum and Minimum Values - 3.1 Execises - Page 212: 48

#### Answer

Absolute maximum: $f(0)=5$ Absolute minimum: $f(-3) = -76$

#### 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 $f(x)=x^3-6x^2+5$ with respect to $x$ keeping $[-3,5]$ as interval. $$\frac{df}{dx}=3x^2-12x$$ $\frac{df}{dx}$ exists on the given interval. Thus, $\frac{df}{dx}=0$, $$3x^2-12x=0\implies x=0, 4.$$ Thus, $f(0)=0^3-6\times 0^2+5=5$. Similarly, $f(4)=-27$. At endpoints, $f(-3)=-76$ & $f(5)=-20$. Thus, absolute maximum $=f(0)=5$ and absolute minimum $=f(-3)=-76$.

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