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