Find the inverse of $y = \frac{x^4-10x^2+13}{5-x^2}$

Question

Find the inverse function of $y = \dfrac{x^4-10x^2+13}{5-x^2}$

I tried to make $x$ the subject as usual but seemed to be stuck. Any idea how to continue?

Answer 1

$\frac{x^4-10 x^2 +13}{5-x^2}=-x^2 +5 + \frac{12}{x^2-5}=y$

$x^2-5=t$

$y=-t + \frac{12}{t}$

$t^2-12+t\cdot y=0$

$t= \frac{-y ± \sqrt{y^2+48}}{2}$

$x^2=5+t =5+ \frac{-y ± \sqrt{y^2+48}}{2}$

$x= ±\sqrt{5+ \frac{-y ± \sqrt{y^2+48}}{2}}$

Answer 2

Hint: Interchange $x$ and $y$: $$x=\frac {y^4-10y^2+13}{5-y^2}$$.

Now solve for $y$. After substitution, use the quadratic formula. Note that $x$ must be restricted to get a one-one function (the original function fails the horizontal line test).