Question

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

Answer 1

`(1)/(2)`

Explanation:

The average rate of change of h(x) in the closed interval [ a, b ] is

`(h(b)-h(a))/(b-a)`

Here [ a , b ] = (- 2, 2 ] , thus

h(b) = h(2) =
`(1)/(8)` (2)³ - 2² = 1 - 4 = - 3

h(a) = h(- 2) =
`(1)/(8)` (- 2)³ - (- 2)² = - 1 - 4 = - 5 , thus

average rate of change =
`(-3-(-5))/(2-(-2))` =
`(2)/(4)` =
`(1)/(2)`