Inverse function of $f(x) = \frac{x+5}{x-2}$

Question

Find the inverse of the function

$f(x) = \frac{x+5}{x-2}$

Here's what I have so far:

$y = \frac{x+5}{x-2}$

$x = \frac{y+5}{y-2}$

$(x)(y-2) = (y+5)$

but this seems to be a dead end.

How should I approach this?

Thank you!

Answer 1

Almost there:

$$x = \frac{y+5}{y-2}$$ $$x(y-2) = y+5$$ $$xy - 2x = y+5$$ $$xy - y = 5 + 2x$$ $$y(x-1) = 5+2x$$

I'll let you finish.

Answer 2

I'm just continuing from where you stopped

$(x)(y-2) = (y+5)$

$xy-2x = y+5$

$xy-y=2x+5$

$y(x-1)=2x+5$

$y=\frac{2x+5}{x-1}$

Answer 3

Hint: Subtract $\displaystyle\frac{x-2}{x-2}$ from the right hand side.

Answer 4

Another approach: your function is a rational function that is a ratio of two linear functions. It's automatic that the inverse will be that same form.

Your function has a vertical asymptote at $x=2$ and a horizontal asymptote at $y=1$, so the inverse function swaps these. So $f^{-1}(x)=\frac{2x+k}{x-1}$ for some $k$. Your function has an $x$-intercept at $-5$, so the inverse has a $y$-intercept at $-5$, which means $k=5$. So $$f^{-1}(x)=\frac{2x+5}{x-1}$$