Question

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

Answer 1

Answer:

$f^{-1}(x)=\dfrac{2}{x+1}-2;$

the domain of the function $f^{-1}(x)$ is   $(-\infty,-1)\cup(-1,\infty);$

the range of the function $f^{-1}(x)$ is   $(-\infty,-2)\cup(-2,\infty).$

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.

$y=\dfrac{2}{x+2}-1,\\ \\\dfrac{2}{x+2}=y+1,\\ \\x+2=\dfrac{2}{y+1},\\ \\x=\dfrac{2}{y+1}-2.$

2. Change x into y and y into x:

$y=\dfrac{2}{x+1}-2,\\ \\f^{-1}(x)=\dfrac{2}{x+1}-2.$

3. The domain of the function $f(x)=\dfrac{2}{x+2}-1$ is $(-\infty,-2)\cup(-2,\infty)$ and the range is   $(-\infty,-1)\cup(-1,\infty).$ Thus, the domain of the function $f^{-1}(x)$ is   $(-\infty,-1)\cup(-1,\infty)$ and the range is  $(-\infty,-2)\cup(-2,\infty)$