h(x)=\frac{1}{8}x^{3}-x^{2}\nover which interval does h have a positive average rate of change?\nchoose 1…

Explanation:

Step1: Recall average rate of change formula

The average rate of change of a function $h(x)$ over the interval $[a,b]$ is $\frac{h(b)-h(a)}{b - a}$.

Step2: Calculate for option A

For $a = 0$, $h(0)=\frac{1}{8}(0)^3-(0)^2 = 0$. For $b = 2$, $h(2)=\frac{1}{8}(2)^3-(2)^2=\frac{8}{8}-4=1 - 4=-3$. Then $\frac{h(2)-h(0)}{2 - 0}=\frac{-3-0}{2}=-\frac{3}{2}<0$.

Step3: Calculate for option B

For $a = 6$, $h(6)=\frac{1}{8}(6)^3-(6)^2=\frac{216}{8}-36 = 27-36=-9$. For $b = 8$, $h(8)=\frac{1}{8}(8)^3-(8)^2=64 - 64=0$. Then $\frac{h(8)-h(6)}{8 - 6}=\frac{0+9}{2}=\frac{9}{2}>0$.

Step4: Calculate for option C

For $a = 0$, $h(0) = 0$. For $b = 6$, $h(6)=-9$. Then $\frac{h(6)-h(0)}{6 - 0}=\frac{-9-0}{6}=-\frac{3}{2}<0$.

Step5: Calculate for option D

For $a = 0$, $h(0)=0$. For $b = 8$, $h(8)=0$. Then $\frac{h(8)-h(0)}{8 - 0}=\frac{0 - 0}{8}=0$.

Answer:

B. $6\leq x\leq8$