what is the average rate of change of the function f(x) = 1/8 x^3 - x^2 over the interval 0, 8? -5/2 -2 -3/2 0

Answer

Explanation:

Step1: Recall the average - rate - of - change formula

The average rate of change of a function $y = f(x)$ over the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 0$, $b = 8$, and $f(x)=\frac{1}{8}x^{3}-x^{2}$.

Step2: Calculate $f(8)$

Substitute $x = 8$ into $f(x)$: [ \begin{align*} f(8)&=\frac{1}{8}\times8^{3}-8^{2}\ &=\frac{1}{8}\times512 - 64\ &=64 - 64\ &=0 \end{align*} ]

Step3: Calculate $f(0)$

Substitute $x = 0$ into $f(x)$: [ f(0)=\frac{1}{8}\times0^{3}-0^{2}=0 ]

Step4: Calculate the average rate of change

[ \begin{align*} \frac{f(8)-f(0)}{8 - 0}&=\frac{0 - 0}{8}\ &=0 \end{align*} ]