# Inverse of f(x)=(x^2+2)/(x-3) ?

May 3, 2018

Depending upon which branch we seek, we have:

${f}^{- 1} \left(x\right) = \frac{x}{2} + \sqrt{{x}^{2} / 4 - 3 x - 2}$

${f}^{- 1} \left(x\right) = \frac{x}{2} - \sqrt{{x}^{2} / 4 - 3 x - 2}$

#### Explanation:

Let:

$y = \frac{{x}^{2} + 2}{x - 3}$

Then in order to find the inverse, ${f}^{- 1} \left(x\right)$, we rearrange the above equations so that we have $x = x \left(y\right)$, that is $x$ as a function of $y$ alone.

So that:

$y \left(x - 3\right) = {x}^{2} + 2$

$\therefore y x - 3 y = {x}^{2} + 2$

$\therefore {x}^{2} - y x + 2 + 3 y = 0$

This is a quadratic in $x$, so we can complete the square:

$\therefore {\left(x - \frac{y}{2}\right)}^{2} - {\left(\frac{y}{2}\right)}^{2} + 2 + 3 y = 0$

$\therefore {\left(x - \frac{y}{2}\right)}^{2} = {y}^{2} / 4 - 3 y - 2$

$\therefore x - \frac{y}{2} = \pm \sqrt{{y}^{2} / 4 - 3 y - 2}$

$\therefore x = \frac{y}{2} \pm \sqrt{{y}^{2} / 4 - 3 y - 2}$

This is not a true function, as it is multi-valued, as a result of the quadratic term in $x$ in the original function.

The graphs of the functions are:

and we clearly see that the inverse is a reflection of the original function in the line $y = x$

So, depending upon which branch we seek, we have:

${f}^{- 1} \left(x\right) = \frac{x}{2} + \sqrt{{x}^{2} / 4 - 3 x - 2}$

${f}^{- 1} \left(x\right) = \frac{x}{2} - \sqrt{{x}^{2} / 4 - 3 x - 2}$