h(x)=x^{2}-1\nover which interval does h have a negative average rate of change?\nchoose 1 answer:\na…

Explanation:

Step1: Recall average - rate - of - change formula

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

Step2: Calculate for option A

For $h(x)=x^{2}-1$ and the interval $[1,4]$, $h(1)=1^{2}-1 = 0$, $h(4)=4^{2}-1=15$. Then $\frac{h(4)-h(1)}{4 - 1}=\frac{15 - 0}{3}=5>0$.

Step3: Calculate for option B

For the interval $[-3,5]$, $h(-3)=(-3)^{2}-1 = 8$, $h(5)=5^{2}-1 = 24$. Then $\frac{h(5)-h(-3)}{5-(-3)}=\frac{24 - 8}{8}=2>0$.

Step4: Calculate for option C

For the interval $[-3,1]$, $h(-3)=(-3)^{2}-1 = 8$, $h(1)=1^{2}-1 = 0$. Then $\frac{h(1)-h(-3)}{1-(-3)}=\frac{0 - 8}{4}=-2<0$.

Step5: Calculate for option D

For the interval $[-1,5]$, $h(-1)=(-1)^{2}-1 = 0$, $h(5)=5^{2}-1 = 24$. Then $\frac{h(5)-h(-1)}{5-(-1)}=\frac{24 - 0}{6}=4>0$.

Answer:

C. $-3\leq x\leq1$