Answer:
The average rate of change of the function [tex]g(x)=x^2+10x+18[/tex] over the interval [tex]-11 \leq x\leq -1[/tex] is -1
Step-by-step explanation:
We are given the function [tex]g(x)=x^2+10x+18[/tex] over the interval [tex]-11 \leq x\leq -1[/tex]
We need to find average rate of change.
The formula used to find average rate of change is : [tex]Average \ rate \ of \ change=\frac{g(b)-g(a)}{b-a}[/tex]
We have b=-1 and a=-11
Finding g(b) = g(-1)
[tex]g(x)=x^2+10x+18\\Putting \ x=-1\\g(-1)=(-1)^2+10(-1)+18\\g(-1)=1-10+18\\g(-1)=9[/tex]
Finding g(a) = g(-11)
[tex]g(x)=x^2+10x+18\\Putting \ x=-11\\g(-11)=(-11)^2+10(-11)+18\\g(-1)=121-110+18\\g(-1)=29[/tex]
Finding average rate of change
[tex]Average \ rate \ of \ change=\frac{g(b)-g(a)}{b-a}\\Average \ rate \ of \ change=\frac{9-29}{-1-(-11)}\\Average \ rate \ of \ change=\frac{-10}{-1+11}\\Average \ rate \ of \ change=\frac{-10}{10}\\Average \ rate \ of \ change=-1[/tex]
So, the average rate of change of the function [tex]g(x)=x^2+10x+18[/tex] over the interval [tex]-11 \leq x\leq -1[/tex] is -1
