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

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Answer 1 · 2018-05-03T21:18:07

Depending upon which branch we seek, we have:

$f^(-1)(x) = x/2 +sqrt(x^2/4-3x-2)$

$f^(-1)(x) = x/2 -sqrt(x^2/4-3x-2)$

Let:

$y = (x^2+2)/(x-3)$

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

So that:

$y(x-3) = x^2+2$

$:. yx-3y = x^2+2$

$:. x^2-yx+2+3y = 0$

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

$:. (x-y/2)^2-(y/2)^2+2+3y = 0$

$:. (x-y/2)^2 = y^2/4-3y-2$

$:. x-y/2 = +-sqrt(y^2/4-3y-2)$

$:. x = y/2 +-sqrt(y^2/4-3y-2)$

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:

![https://www.wolframalpha.com/input/?i=inverse+y%3D(x%5E2%2B2)%2F(x-3)](https://d2jmvrsizmvf4x.cloudfront.net/zgaQs9szQ2GLl2J1CwFC_gif%26s%3D26)

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:

$f^(-1)(x) = x/2 +sqrt(x^2/4-3x-2)$

$f^(-1)(x) = x/2 -sqrt(x^2/4-3x-2)$

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