Question

hw 20 - absolute extrema section 3.6: problem 5
(1 point)
find the extreme values of the function ( f ) on the interval ( 0.6,4 ). if an extreme value does not exist, enter dne.
( f(x)=x^{2}+\frac{2}{x} )
absolute minimum value:
absolute maximum value:
(round to three decimal places as needed.)

Explanation:

Step1: Find the derivative of \(f(x)\)

Using the power rule, if \(f(x)=x^{2}+\frac{2}{x}=x^{2}+2x^{- 1}\), then \(f^{\prime}(x)=2x - 2x^{-2}=2x-\frac{2}{x^{2}}=\frac{2x^{3}-2}{x^{2}}\).

Step2: Find the critical points

Set \(f^{\prime}(x) = 0\), so \(\frac{2x^{3}-2}{x^{2}}=0\). Since \(x^{2}
eq0\) (for \(x\in[0.6,4]\)), then \(2x^{3}-2 = 0\).

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Step3: Evaluate the function at the critical point and endpoints

• Evaluate \(f(x)\) at \(x = 0.6\): \(f(0.6)=(0.6)^{2}+\frac{2}{0.6}=0.36+\frac{10}{3}\approx0.36 + 3.333=3.693\).
• Evaluate \(f(x)\) at \(x = 1\): \(f(1)=1^{2}+\frac{2}{1}=1 + 2=3\).
• Evaluate \(f(x)\) at \(x = 4\): \(f(4)=4^{2}+\frac{2}{4}=16+0.5=16.5\).

Answer:

Absolute minimum value: \(3.000\)
Absolute maximum value: \(16.500\)