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pantera1 [17]
3 years ago
13

h(x)= \dfrac{1}{8} x^3 - x^2h(x)= 8 1 ​ x 3 −x 2 h, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided b

y, 8, end fraction, x, cubed, minus, x, squared What is the average rate of change of hhh over the interval -2\leq x\leq 2−2≤x≤2minus, 2, is less than or equal to, x, is less than or equal to, 2?
Mathematics
1 answer:
Nesterboy [21]3 years ago
6 0

Answer:

\dfrac{1}{2}

Step-by-step explanation:

Given the function h(x)=\dfrac{1}{8}x^3-x^2

The average rate of change of the function h(x) over the interval a\le x\le b can be calculated using formula

\dfrac{h(b)-h(a)}{b-a}

In your case,

a=-2,\\ \\b=2,

so the average rate of change is

\dfrac{h(2)-h(-2)}{2-(-2)}\\ \\=\dfrac{(\frac{1}{8}\cdot 2^3-2^2)-(\frac{1}{8}\cdot (-2)^3-(-2)^2)}{4}\\ \\=\dfrac{(1-4)-(-1-4)}{4}\\ \\=\dfrac{-3+5}{4}\\ \\=\dfrac{2}{4}\\ \\=\dfrac{1}{2}

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