Question

b. $h(x)=cos(5^{circ}+7)$

1. use the table below to find the average rate of change in $fcirc g$ on $-1,2$

$x$	-3	-2	-1	0	1	2	3
$f(x)$	11	9	7	5	3	1	-1
$g(x)$	-8	-3	0	1	0	-3	-8

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = F(x)$ on the interval $[a,b]$ is $\frac{F(b)-F(a)}{b - a}$. Here $F(x)=(f\circ g)(x)$ and $a=-1$, $b = 2$.

Step2: Find $(f\circ g)(-1)$ and $(f\circ g)(2)$

First, when $x=-1$, $g(-1)=0$. Then $(f\circ g)(-1)=f(g(-1))=f(0)$. From the table, $f(0)=5$.
When $x = 2$, $g(2)=-3$. Then $(f\circ g)(2)=f(g(2))=f(-3)$. From the table, $f(-3)=11$.

Step3: Calculate the average rate of change

The average rate of change of $f\circ g$ on $[-1,2]$ is $\frac{(f\circ g)(2)-(f\circ g)(-1)}{2-(-1)}=\frac{f(g(2))-f(g(-1))}{3}=\frac{11 - 5}{3}$.
$\frac{11 - 5}{3}=\frac{6}{3}=2$.

Answer:

$2$