Question

find the absolute maximum and minimum values of the following function on the given interval. then graph the function. identify the points on the graph where the absolute extrema occur.
h(x)=2sqrt3{x}, - 1leq xleq8
find the absolute maximum. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the absolute maximum value occurs at x =
(use a comma to separate answers as needed.)
b. there is no absolute maximum.

Explanation:

Step1: Rewrite the function

Rewrite $h(x)=2\sqrt[3]{x}$ as $h(x) = 2x^{\frac{1}{3}}$.

Step2: Find the derivative

Using the power - rule $(x^n)'=nx^{n - 1}$, we have $h'(x)=\frac{2}{3}x^{-\frac{2}{3}}=\frac{2}{3\sqrt[3]{x^{2}}}$.

Step3: Find critical points

Set $h'(x) = 0$. Since the numerator of $h'(x)$ is non - zero ($2
eq0$), there are no values of $x$ for which $h'(x)=0$. But $h'(x)$ is undefined at $x = 0$. So $x = 0$ is a critical point.

Step4: Evaluate the function at critical and end - points

Evaluate $h(x)$ at $x=-1$, $x = 0$, and $x = 8$.
When $x=-1$, $h(-1)=2\sqrt[3]{-1}=-2$.
When $x = 0$, $h(0)=2\sqrt[3]{0}=0$.
When $x = 8$, $h(8)=2\sqrt[3]{8}=2\times2 = 4$.

Answer:

A. The absolute maximum value 4 occurs at x = 8.