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}, -1≤x≤8
a. the absolute minimum value - 2 occurs at x = - 1. (use a comma to separate answers as needed.)
b. there is no absolute minimum.
choose the correct graph of the function.

Explanation:

Step1: Evaluate function at endpoints

First, find $h(-1)$ and $h(8)$. For $h(x)=2\sqrt[3]{x}$, when $x = - 1$, $h(-1)=2\sqrt[3]{-1}=2\times(-1)=-2$. When $x = 8$, $h(8)=2\sqrt[3]{8}=2\times2 = 4$.

Step2: Analyze critical - points

The derivative of $h(x)=2x^{\frac{1}{3}}$ is $h^\prime(x)=\frac{2}{3x^{\frac{2}{3}}}$. The derivative is undefined at $x = 0$, but it exists for all $x
eq0$ in the domain. Since the function is continuous on $[-1,8]$, we consider the values at the endpoints $x=-1$ and $x = 8$.

Step3: Determine absolute extrema

Comparing $h(-1)=-2$ and $h(8)=4$, the absolute minimum value is $-2$ at $x=-1$ and the absolute maximum value is $4$ at $x = 8$.

Answer:

A. The absolute minimum value $-2$ occurs at $x=-1$.
The absolute maximum value $4$ occurs at $x = 8$.
(For the graph - related part, without seeing the full details of the graphs, we can note that the function $y = 2\sqrt[3]{x}$ is a cube - root function. It passes through the points $(-1,-2)$ and $(8,4)$. The general shape of the cube - root function $y=\sqrt[3]{x}$ is a smooth curve that increases from negative infinity to positive infinity, and $y = 2\sqrt[3]{x}$ is a vertical stretch of $y=\sqrt[3]{x}$ by a factor of 2. Based on the information about the extrema, the correct graph should have a point at $(-1,-2)$ as the absolute minimum and $(8,4)$ as the absolute maximum within the interval $[-1,8]$.)