Answer
a. $19$
b. $1$
Work Step by Step
a. $f(x)=x^3+1 , [2,3]$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$$
$f(x_{2})=f(3)=3^{3}+1$
$f(3)=28$
$f(x_{1})=f(2)=2^{3}+1$
$f(2)=9$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{f(3)-f(2)}{x_{2}-x_{1}}$$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{28-9}{3-2}$$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{19}{1}$$
$$\frac{\Delta{y}}{\Delta{x}}=19$$
b. $f(x)=x^3+1, [-1,1]$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$$
$f(x_{2})=f(1)=1^{3}+1$
$f(1)=2$
$f(x_{1})=f(-1)=-1^{3}+1$
$f(-1)=0$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{f(1)-f(-1)}{x_{2}-x_{1}}$$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{2-0}{1-(-1)}$$
$$\frac{\Delta{y}}{\Delta{x}}=\frac{2}{2}$$
$$\frac{\Delta{y}}{\Delta{x}}=1$$
