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EXAMPLE 3Finding Absolute Maximum and Minimum Values

Find the absolute maximum value and the absolute minimum value of each function:

$f(x)=x^{3}-6x^{2}+9x+2$ on $[0,2]$
$g(x)= \dfrac{(x-2)^{2/3}}{x}$ on $[1,10]$

(a) The function $f$ , a polynomial function, is continuous on the closed interval $[0,2]$ , so the Extreme Value Theorem guarantees that $f$ has an absolute maximum value and an absolute minimum value on the interval. We follow the steps for finding the absolute extreme values to identify them.

Step 1 From Example 2(a), the critical numbers of $f$ are $1$ and $3$ . We exclude $3$ , since it is not in the interval $(0,2)$ .

Step 2 Find the value of $f$ at the critical number $1$ and at the endpoints $0$ and $2$ : $f(1)=6\qquad f( 0) =2\qquad f(2) =4$

Step 3 The largest value $6$ is the absolute maximum value of $f$ ; the smallest value $2$ is the absolute minimum value of $f$ .

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(b) The function $g$ is continuous on the closed interval $[1,10]$ , so $g$ has an absolute maximum and an absolute minimum on the interval.

Step 1 From Example 2(c), the critical numbers of $g$ are $2$ and $6$ . Both critical numbers are in the interval $(1,10)$ .

Step 2 We evaluate $g$ at the critical numbers $2$ and $6$ and at the endpoints $1$ and $10$ :

$x$	${\dfrac{( x-2) ^{2/3}}{x}}$	${g(x)}$
$1$	$\dfrac{(1-2) ^{2/3}}{1}=(-1) ^{2/3}$	$1$	$\longleftarrow$ absolute maximum value
$2$	$\dfrac{( 2-2) ^{2/3}}{2}$	$0$	$\longleftarrow$ absolute minimum value
$6$	$\dfrac{(6-2) ^{2/3}}{6}=\dfrac{4^{2/3}}{6}$	$\approx\! 0.42$
$10$	$\dfrac{(10-2) ^{2/3}}{10}=\dfrac{8^{2/3}}{10}=\dfrac{4}{10}$	$0.4$

Step 3 The largest value $1$ is the absolute maximum value; the smallest value $0$ is the absolute minimum value.