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

Explanation:

I will assume you meant
`h(x)=(1)/(8^(3))-x^2`, which is equal to
`h(x)=(1)/(512)-x^2`.

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