Inverse Function That Includes Fraction

Question

Nice inverse function I am struggling on, totally forgotten how to move the fraction over:

$$g(x)= \frac 1 {x-2}+5 \qquad (x>2)$$

Find the inverse function $g^{-1}$ specifying the rule domain and image set.

Any help is appreciated!

Answer 1

Note that $$g(x)= \frac 1 {x-2}+5 = y \Rightarrow \frac 1 {x-2} = y -5\\x - 2 = \frac{1}{y-5}\\x = \frac{1}{y-5} + 2$$ hence $$g^{-1}(y) = \dfrac 1{y-5} + 2$$ Now $$x > 2 \Rightarrow \frac{1}{y-5} > 0 \Rightarrow y > 5$$

Answer 2

$\newcommand{\leftlong}{\longleftarrow\!\shortmid}$ $$ x \overset{\text{subtract 2}}\mapsto x-2 \overset{\text{take reciprocal}}\mapsto \frac 1 {x-2} \overset{\text{add 5}}\mapsto \frac 1 {x-2} + 5 $$

What gets done last gets undone first:

$$ \frac 1 {y-5} + 2\ \overset{\text{add 2}}\leftlong\ \frac 1 {y-5}\ \overset{\text{take reciprocal}} \leftlong\ y-5\ \overset{\text{subtract 5}} \leftlong\ y $$

So $$g^{-1}(y) = \dfrac 1{y-5} + 2.$$

(If the variable $x$ were to appear twice or more in the expression defining the function, then this technique would not work and you'd have to resort to algebra.)