Q:

Algebra 2! Help! I WILL GIVE BRAINLIEST!!!!

Accepted Solution

A:
Answer:[tex]f^{-1}(x)=\dfrac{2}{x+1}-2;[/tex]the domain of the function [tex]f^{-1}(x)[/tex] is   [tex](-\infty,-1)\cup(-1,\infty);[/tex]the range of the function [tex]f^{-1}(x)[/tex] is   [tex](-\infty,-2)\cup(-2,\infty).[/tex]Step-by-step explanation:To write the rule of inverse function you need to express x in terms of y and then change x into y and y into x.1. [tex]y=\dfrac{2}{x+2}-1,\\ \\\dfrac{2}{x+2}=y+1,\\ \\x+2=\dfrac{2}{y+1},\\ \\x=\dfrac{2}{y+1}-2.[/tex]2. Change x into y and y into x:[tex]y=\dfrac{2}{x+1}-2,\\ \\f^{-1}(x)=\dfrac{2}{x+1}-2.[/tex]3. The domain of the function [tex]f(x)=\dfrac{2}{x+2}-1[/tex] is [tex](-\infty,-2)\cup(-2,\infty)[/tex] and the range is   [tex](-\infty,-1)\cup(-1,\infty).[/tex] Thus, the domain of the function [tex]f^{-1}(x)[/tex] is   [tex](-\infty,-1)\cup(-1,\infty)[/tex] and the range is  [tex](-\infty,-2)\cup(-2,\infty)[/tex]