find the absolute maximum and minimum values of the function over the indicated interval, and indicate the x…

Explanation:

Step1: Find the derivative

Use the chain - rule. If $y = u^{\frac{2}{3}}$ and $u=x + 5$, then $\frac{dy}{du}=\frac{2}{3}u^{-\frac{1}{3}}$ and $\frac{du}{dx}=1$. So $f^\prime(x)=\frac{2}{3}(x + 5)^{-\frac{1}{3}}=\frac{2}{3\sqrt[3]{x + 5}}$.

Step2: Find the critical points

Set $f^\prime(x)=0$. Since the numerator of $f^\prime(x)$ is non - zero ($2\neq0$), $f^\prime(x)$ has no solution for $f^\prime(x) = 0$. But $f^\prime(x)$ is undefined when $x+5 = 0$, i.e., $x=-5$.

Step3: Evaluate the function at critical and endpoints

Evaluate $f(x)$ at $x=-8,x=-5,x = 7$. For $x=-8$: $f(-8)=(-8 + 5)^{\frac{2}{3}}-3=(-3)^{\frac{2}{3}}-3=\sqrt[3]{9}-3$. For $x=-5$: $f(-5)=(-5 + 5)^{\frac{2}{3}}-3=-3$. For $x = 7$: $f(7)=(7 + 5)^{\frac{2}{3}}-3=12^{\frac{2}{3}}-3=\sqrt[3]{144}-3$.

Step4: Compare the values

We know that $\sqrt[3]{9}-3\approx2.08-3=-0.92$, $-3$, $\sqrt[3]{144}-3\approx5.24-3 = 2.24$.

Answer:

The absolute maximum value is $\sqrt[3]{144}-3\approx2.24$ and it occurs at $x = 7$. The absolute minimum value is $-3$ and it occurs at $x=-5$.