find the absolute maximum and minimum values of the function, if they exist, over the indicated interval…

Explanation:

Step1: Find the derivative

Differentiate $f(x)=x^{2}+\frac{80}{x}=x^{2}+80x^{-1}$ using the power - rule. The derivative $f^\prime(x)=2x - 80x^{-2}=2x-\frac{80}{x^{2}}=\frac{2x^{3}-80}{x^{2}}$.

Step2: Set the derivative equal to zero

Set $f^\prime(x) = 0$, so $\frac{2x^{3}-80}{x^{2}}=0$. Since the denominator $x^{2}\neq0$ for $x\in(0,\infty)$, we solve $2x^{3}-80 = 0$. Then $x^{3}=40$, and $x=\sqrt[3]{40}$.

Step3: Use the second - derivative test

Differentiate $f^\prime(x)$ to get the second - derivative. $f^{\prime\prime}(x)=2 + 160x^{-3}=2+\frac{160}{x^{3}}$. Substitute $x = \sqrt[3]{40}$ into $f^{\prime\prime}(x)$: $f^{\prime\prime}(\sqrt[3]{40})=2+\frac{160}{40}=2 + 4=6>0$. So $f(x)$ has a local minimum at $x=\sqrt[3]{40}$.

Step4: Evaluate the function at the critical point

$f(\sqrt[3]{40})=(\sqrt[3]{40})^{2}+\frac{80}{\sqrt[3]{40}}=40^{\frac{2}{3}}+80\times40^{-\frac{1}{3}}=40^{\frac{2}{3}} + 2\times40^{\frac{2}{3}}=3\times40^{\frac{2}{3}}$. As $x\to0^{+}$, $f(x)\to\infty$, and as $x\to\infty$, $f(x)\to\infty$.

Answer:

The absolute minimum value is $3\times40^{\frac{2}{3}}\approx3\times7.37=22.11$ and it occurs at $x = \sqrt[3]{40}\approx3.42$. There is no absolute maximum value over the interval $(0,\infty)$.