f(x)=x^{3}-9x\nover which interval does f have a positive average rate of change?\nchoose 1 answer:\na…

Explanation:

Step1: Recall average - rate - of - change formula

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

Step2: Calculate for option A

For $f(x)=x^{3}-9x$ and the interval $[-2,1]$, $a=-2$, $b = 1$. $f(-2)=(-2)^{3}-9\times(-2)=-8 + 18=10$, $f(1)=1^{3}-9\times1=1 - 9=-8$. The average rate of change is $\frac{f(1)-f(-2)}{1-(-2)}=\frac{-8 - 10}{3}=\frac{-18}{3}=-6$.

Step3: Calculate for option B

For the interval $[-4,-1]$, $a=-4$, $b=-1$. $f(-4)=(-4)^{3}-9\times(-4)=-64 + 36=-28$, $f(-1)=(-1)^{3}-9\times(-1)=-1 + 9=8$. The average rate of change is $\frac{f(-1)-f(-4)}{-1-(-4)}=\frac{8+28}{3}=\frac{36}{3}=12$.

Step4: Calculate for option C

For the interval $[-1,2]$, $a=-1$, $b = 2$. $f(-1)=-1 + 9=8$, $f(2)=2^{3}-9\times2=8 - 18=-10$. The average rate of change is $\frac{f(2)-f(-1)}{2-(-1)}=\frac{-10 - 8}{3}=\frac{-18}{3}=-6$.

Step5: Calculate for option D

For the interval $[-3,3]$, $a=-3$, $b = 3$. $f(-3)=(-3)^{3}-9\times(-3)=-27+27 = 0$, $f(3)=3^{3}-9\times3=27 - 27=0$. The average rate of change is $\frac{f(3)-f(-3)}{3-(-3)}=\frac{0 - 0}{6}=0$.

Answer:

B. $[-4,-1]$