Question

H(x)=

Answer 1

The average rate of change of the function over −2 ≤ x ≤ 2 is 1/2

How to determine the average rate of change?

The function is given as:

$h(x) = \frac 18x^3 - x^2$

The interval is given as:

−2 ≤ x ≤ 2

Calculate h(2) and h(-2)

$h(2) = \frac 18 * 2^3 - (2)^2$

h(2) = -3

$h(-2) = \frac 18 * (-2)^3 - (-2)^2$

h(-2) = -5

The average rate of change is then calculated as:

$m = \frac{h(-2) - h(2)}{-2-2}$

This gives

$m = \frac{-5 + 3}{-4}$

Evaluate

m = 1/2

Hence, the average rate of change of the function over −2 ≤ x ≤ 2 is 1/2

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Complete question

$h(x) = \frac 18x^3 - x^2$

What is the average rate of change of h over the interval −2 ≤ x ≤ 2