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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 by, 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?

Respuesta :

frika

Answer:

[tex]\dfrac{1}{2}[/tex]

Step-by-step explanation:

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

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

[tex]\dfrac{h(b)-h(a)}{b-a}[/tex]

In your case,

[tex]a=-2,\\ \\b=2,[/tex]

so the average rate of change is

[tex]\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}[/tex]

Answer:

I... don't understand why the answer below had such low ratings??!

The answer 1/2 is correct on khan academy if Im looking that the right question..

Step-by-step explanation:

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