h(x)=\frac{1}{8}x^{3}-x^{2}
what is the average rate of change of h over the interval -2\leq x\leq2?

Question

Accepted Answer

# Explanation:
## Step1: Recall the formula for average rate of change
The average rate of change of a function $y = h(x)$ over the interval $[a,b]$ is $\frac{h(b)-h(a)}{b - a}$. Here $a=-2$, $b = 2$.
## Step2: Calculate $h(-2)$
Substitute $x=-2$ into $h(x)=\frac{1}{8}x^{3}-x^{2}$:
$$
\begin{align*}
h(-2)&=\frac{1}{8}\times(-2)^{3}-(-2)^{2}\
&=\frac{1}{8}\times(-8)-4\
&=-1 - 4\
&=-5
\end{align*}
$$
## Step3: Calculate $h(2)$
Substitute $x = 2$ into $h(x)=\frac{1}{8}x^{3}-x^{2}$:
$$
\begin{align*}
h(2)&=\frac{1}{8}\times2^{3}-2^{2}\
&=\frac{1}{8}\times8-4\
&=1-4\
&=-3
\end{align*}
$$
## Step4: Calculate the average rate of change
Use the formula $\frac{h(b)-h(a)}{b - a}$, with $a=-2$, $b = 2$, $h(a)=-5$, $h(b)=-3$:
$$
\begin{align*}
\frac{h(2)-h(-2)}{2-(-2)}&=\frac{-3-(-5)}{2 + 2}\
&=\frac{-3 + 5}{4}\
&=\frac{2}{4}\
&=\frac{1}{2}
\end{align*}
$$

# 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: