How to find the inverse of $f(x) = \frac{x+2}x$?

Question

What approach would be ideal in finding the inverse of $f(x) = \frac{x+2}x$?

Answer 1

The idea of an inverse is reflecting across the line of $y=x$. To obtain the inverse, all you have to do is flip all the $x$'s with $y$ (or $f(x)$) and vice versa. We have:

$$f(x)=\frac{x+2}{x}$$

Applying the rules I just told you, the new function with the inverse ($f^{-1}(x)$) will be: $$x=\frac{f^{-1}(x)+2}{f^{-1}(x)}$$ Now simplify so you can isolate for $f^{-1}(x)$: $$x=\frac{f^{-1}(x)}{f^{-1}(x)}+\frac2{f^{-1}(x)} \\ \Rightarrow x-1=\frac2{f^{-1}(x)} \\ \therefore f^{-1}(x)=\frac2{x-1}$$

Answer 2

$$f(x)=\dfrac{x+2}{x}\\ \implies x\times \left(f(x)-1\right)=2\\ \implies x=?$$ The expression $?=f^{-1}(x)$.

Answer 3

First of all the domain for f is $\mathbb{R}\setminus\{0\}$. $y=\frac{x+2}{x}$ give us $xy=x+2$ so $x(y-1)=2$ and for $y\neq 1$ we have: $x=\frac{2}{y-1}$. Hence $g(x)=\frac{2}{x-1}$ is the inverse of $f$ with domain $\mathbb{R}\setminus\{1\}$. Verify that $f\circ g= id=g\circ f$.

Answer 4

Note that $\frac{x+2}{x}=1+\frac2x$. You should be able to invert this by inspection: $$y = 1+\frac2x \iff x= \frac{2}{y-1}$$ Thus $$\boxed{f^{-1}(y) =\dfrac{2}{y-1}}$$ for $y\neq 1$.

Answer 5

For a more rigorous proof, let $g(x)=\frac{2}{x-1}$ for $x\neq 1$ (however you may obtain it is irrelevant to the proof), and verify that, as user Test123 pointed out, $f\circ g= id=g\circ f$. Only then can you conclude that $g$ is the inverse of $f$