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H(x) = 1/8 x^3 -x2 what is the average rate of change of h over the Interval-2 < x< 2

User Danie A
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The average rate of change of a function over an interval is the total change in the function's output (or y-value) divided by the total change in the function's input (or x-value) over that interval.

Given the function h(x) = 1/8 x^3 - x^2, the average rate of change over the interval -2 < x < 2 is:

(h(2) - h(-2)) / (2 - (-2))

First, we have to find h(2) and h(-2) by substituting these values in the function:

h(2) = 1/8 (2)^3 - (2)^2 = 1/8 * 8 - 4 = 0.5

h(-2) = 1/8 (-2)^3 - (-2)^2 = 1/8 * -8 - 4 = -4.5

So, the average rate of change is:

(0.5 - (-4.5)) / (2 - (-2)) = 5 / 4 = 1.25

Therefore, the average rate of change of h over the interval -2 < x < 2 is 1.25

User Atasha
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