Question

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?

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:

\(\frac{1}{2}\)