h(x) = \frac{1}{8}x^3 - x^2\
what is the average rate of change of h over the interval -2 \leq x \leq 2?

Question

Accepted Answer

# Explanation:
## Step1: Recall the formula for average rate of change
The average rate of change of a function $ h(x) $ over the interval $[a, b]$ is given by $\frac{h(b) - h(a)}{b - a}$. Here, $ a=-2 $, $ b = 2 $.

## Step2: Calculate $ h(-2) $
Substitute $ x=-2 $ into $ h(x)=\frac{1}{8}x^{3}-x^{2} $:
$$
\begin{align*}
h(-2)&=\frac{1}{8}(-2)^{3}-(-2)^{2}\
&=\frac{1}{8}(-8)-4\
&=- 1-4\
&=-5
\end{align*}
$$

## Step3: Calculate $ h(2) $
Substitute $ x = 2 $ into $ h(x)=\frac{1}{8}x^{3}-x^{2} $:
$$
\begin{align*}
h(2)&=\frac{1}{8}(2)^{3}-(2)^{2}\
&=\frac{1}{8}(8)-4\
&=1 - 4\
&=-3
\end{align*}
$$

## Step4: Calculate the average rate of change
Using the formula $\frac{h(b)-h(a)}{b - a}$ with $ a=-2 $, $ b = 2 $, $ h(-2)=-5 $, $ h(2)=-3 $:
$$
\begin{align*}
\text{Average rate of change}&=\frac{h(2)-h(-2)}{2-(-2)}\
&=\frac{-3-(-5)}{2 + 2}\
&=\frac{-3 + 5}{4}\
&=\frac{2}{4}\
&=\frac{1}{2}
\end{align*}
$$

# Answer:
$\frac{1}{2}$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the formula for average rate of change

Step2: Calculate \( h(-2) \)

Step3: Calculate \( h(2) \)

Step4: Calculate the average rate of change

Answer: