Answer
a. $2$
b. $0$
Work Step by Step
a. $g(x)=x^2-2x , [1,3]$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{g(x_{2})-g(x_{1})}{x_{2}-x_{1}}$$
$g(x_{2})=g(3)=3^{2}-2(3)$
$g(3)=3$
$g(x_{1})=g(1)=1^{2}-2(1)$
$g(1)=-1$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{g(3)-g(1)}{x_{2}-x_{1}}$$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{3-(-1)}{3-1}$$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{4}{2}$$
$$\frac{\Delta{y}}{\Delta{x}}=2$$
b. $g(x)=x^2-2x , [-2,4]$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{g(x_{2})-g(x_{1})}{x_{2}-x_{1}}$$
$g(x_{2})=g(4)=4^{2}-2(4)$
$g(4)=8$
$g(x_{1})=g(-2)=(-2)^{2}-2(-2)$
$g(-2)=8$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{g(4)-g(-2)}{x_{2}-x_{1}}$$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{8-8}{4-(-2)}$$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{0}{6}$$
$$\frac{\Delta{y}}{\Delta{x}}=0$$
