Question

h(x) = 1/8^3 -x^2 What is the average rate of change of h over the interval -2 ≤ x ≤ 2?​

Answer 1

Answer: 0

Step-by-step explanation:

I will assume you meant $h(x)=\frac{1}{8^{3}}-x^2$, which is equal to $h(x)=\frac{1}{512}-x^2$.

$f(-2)=\frac{1}{512}-(-2)^{2}=-\frac{2047}{512}\\\\f(2)=\frac{1}{512}-2^{2}=-\frac{2047}{512}\\\\\therefore \frac{f(2)-f(-2)}{2-(-2)}=\frac{-\frac{2047}{512}-\left(-\frac{2047}{512} \right)}{2-(-2)}=\boxed{0}$