13. find the absolute maximum and minimum of $f(x)=x^{2}+\frac{2}{x}$ on $\frac{1}{2},2$.

Explanation:

Step1: Find the derivative

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

Step2: Find the critical points

Set $f^\prime(x) = 0$, so $\frac{2x^{3}-2}{x^{2}}=0$. Since the denominator $x^{2}\neq0$ for the domain we care about, we solve $2x^{3}-2 = 0$. Then $x^{3}-1=0$, which gives $x = 1$.

Step3: Evaluate the function at critical points and endpoints

Evaluate $f(x)$ at $x=\frac{1}{2},1,2$. When $x=\frac{1}{2}$, $f(\frac{1}{2})=(\frac{1}{2})^{2}+\frac{2}{\frac{1}{2}}=\frac{1}{4}+4=\frac{1 + 16}{4}=\frac{17}{4}$. When $x = 1$, $f(1)=1^{2}+\frac{2}{1}=1 + 2=3$. When $x = 2$, $f(2)=2^{2}+\frac{2}{2}=4 + 1=5$.

Answer:

The absolute minimum value of $f(x)$ on $[\frac{1}{2},2]$ is $3$ at $x = 1$, and the absolute maximum value is $\frac{17}{4}$ at $x=\frac{1}{2}$.