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

Answer:

$\frac{1}{2}$

Step-by-step explanation:

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

$\frac{h(b)-h(a)}{b-a}$

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

h(b) = h(2) = $\frac{1}{8}$ (2)³ - 2² = 1 - 4 = - 3

h(a) = h(- 2) = $\frac{1}{8}$ (- 2)³ - (- 2)² = - 1 - 4 = - 5 , thus

average rate of change = $\frac{-3-(-5)}{2-(-2)}$ = $\frac{2}{4}$ = $\frac{1}{2}$