We have been given a function [tex]h(x)=\frac{1}{8}x^3-x^2[/tex]. We are asked to find the average rate of change of our given function over the interval [tex]-2\leq x\leq 2[/tex].
We will use average rate of change formula to solve our given problem.
[tex]\text{Average rate of change}=\frac{f(b)-f(a)}{b-a}[/tex]
[tex]\text{Average rate of change}=\frac{h(2)-h(-2)}{2-(-2)}[/tex]
[tex]\text{Average rate of change}=\frac{\frac{1}{8}(2)^3-2^2-(\frac{1}{8}(-2)^3-(-2)^2)}{2+2}[/tex]
[tex]\text{Average rate of change}=\frac{\frac{1}{8}(8)-4-(\frac{1}{8}(-8)-(4))}{4}[/tex]
[tex]\text{Average rate of change}=\frac{1-4-(-1-4)}{4}[/tex]
[tex]\text{Average rate of change}=\frac{-3-(-5)}{4}[/tex]
[tex]\text{Average rate of change}=\frac{-3+5}{4}[/tex]
[tex]\text{Average rate of change}=\frac{2}{4}[/tex]
[tex]\text{Average rate of change}=\frac{1}{2}[/tex]
Therefore, the average rate of change over the interval [tex]-2\leq x\leq 2[/tex] is [tex]\frac{1}{2}[/tex].
