Question

on which interval does the function g(x)=4x - 3x^2 have an average rate of change equal to
x = - 4 to x = 2
x = - 4 to x = 0
x = - 4 to x = 3
x = - 4 to x = 1

Explanation:

Step1: Recall average rate of change formula

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

Step2: Calculate $g(x)$ at endpoints for each interval

For the interval $x=-4$ to $x = 0$:
$a=-4$, $b = 0$.
$g(-4)=4\times(-4)-3\times(-4)^2=-16-3\times16=-16 - 48=-64$.
$g(0)=4\times0-3\times0^2 = 0$.
The average rate of change is $\frac{g(0)-g(-4)}{0-(-4)}=\frac{0 - (-64)}{4}=\frac{64}{4}=16$.

For the interval $x=-4$ to $x = 1$:
$a=-4$, $b = 1$.
$g(-4)=-64$ (calculated above), $g(1)=4\times1-3\times1^2=4 - 3=1$.
The average rate of change is $\frac{g(1)-g(-4)}{1-(-4)}=\frac{1-(-64)}{5}=\frac{65}{5}=13$.

For the interval $x=-4$ to $x = 2$:
$a=-4$, $b = 2$.
$g(-4)=-64$, $g(2)=4\times2-3\times2^2=8 - 12=-4$.
The average rate of change is $\frac{g(2)-g(-4)}{2-(-4)}=\frac{-4-(-64)}{6}=\frac{60}{6}=10$.

For the interval $x=-4$ to $x = 3$:
$a=-4$, $b = 3$.
$g(-4)=-64$, $g(3)=4\times3-3\times3^2=12 - 27=-15$.
The average rate of change is $\frac{g(3)-g(-4)}{3-(-4)}=\frac{-15-(-64)}{7}=\frac{49}{7}=7$.

Answer:

The interval $x=-4$ to $x = 0$ has an average rate of change equal to 16.