Respuesta :
the average rate of change of h is 1/2 .
Step-by-step explanation:
Here we have , a function h(x)=1/8x^3-x^2 or , [tex]h(x)=\frac{1}{8}x^3-x^2[/tex] . We need to find rate of change of function over [tex]-2<x<2[/tex] . Let's find out:
We know that , Rate of change of function is :
[tex]\frac{f(b)-f(a)}{b-a}[/tex]
According to question we have ,
⇒ [tex]\frac{f(-2)-f(2)}{-2-2}[/tex]
[tex]f(-2) = \frac{1}{8}(-2)^3-(-2)^2 = \frac{1}{8}(-8)-4 = -5\\f(2) = \frac{1}{8}(2)^3-(2)^2 = \frac{1}{8}(8)-4 = -3[/tex]
⇒ [tex]\frac{-5-(-3)}{-2-2}[/tex]
⇒ [tex]\frac{-2}{-4}[/tex]
⇒ [tex]\frac{1}{2}[/tex]
Therefore , the average rate of change of h is 1/2 .
