Question

h(x)=\frac{1}{8}x^{3}-x^{2}
what is the average rate of change of h over the interval - 2leq xleq2?

Explanation:

Step1: Recall average - rate - of - change formula

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 $h(x)=\frac{1}{8}x^{3}-x^{2}$:
$h(2)=\frac{1}{8}\times2^{3}-2^{2}=\frac{1}{8}\times8 - 4=1 - 4=-3$.

Step3: Calculate $h(-2)$

Substitute $x=-2$ into $h(x)=\frac{1}{8}x^{3}-x^{2}$:
$h(-2)=\frac{1}{8}\times(-2)^{3}-(-2)^{2}=\frac{1}{8}\times(-8)-4=-1 - 4=-5$.

Step4: Calculate the average rate of change

$\frac{h(2)-h(-2)}{2-(-2)}=\frac{-3-(-5)}{2 + 2}=\frac{-3 + 5}{4}=\frac{2}{4}=\frac{1}{2}$.

Answer:

$\frac{1}{2}$