Inverse of fourth root function

Question

I have a function in the form $$y=a\sqrt[^4]{b^2-(x-c)^2}, \qquad a, b, c\in \mathbb{R}$$

but I'm having trouble finding its inverse. That is, solving for $x$. The solution should seem pretty trivial with putting both sides to the $4^{\text{th}}$ power, rearranging, and applying the quadratic formula, but I can't seem to obtain the correct solution. Any suggestions? Thanks!

Answer 1

Taking the forth power, you get $$ y^4 = a^4 (b^2-(x-c)^2) \Leftrightarrow \frac{y^4}{a^4}-b^2 = -(x-c)^2 \Leftrightarrow (x-c)^2 = b^2-\frac{y^4}{a^4} $$

so, finally

$$ x = c \pm \sqrt{b^2-\frac{y^4}{a^4}}. $$

We do know that $a \ne 0$ (if it was zero, then $y=0$ and there would be no inverse!), but now you must discuss how to pick the $\pm$ sign and what does that say about the existence of the inverse in appropriate intervals.

If look into it, you'll see that you must have $x \in [c-b, c+b]$ and that $y(x)$ is invertible on $[c-b,c]$ and on $[c,c+b]$, but not on the full interval. Hence the choice of the sign.

Answer 2

\begin{align} y=a\sqrt[^4]{b^2-(x-c)^2} &\implies y^{4} = a^{4}(b^{2} - (x-c)^2)\\ &\implies \frac{y^{4}}{a^{4}} = b^{2} - x^{2}+2cx-c^{2}\\ &\implies x^{2} - 2cx + \left(\frac{y^{4}}{a^{4}}-b^{2}+c^{2}\right) = 0\\ &\implies x = \frac{2c \pm \sqrt{4c^2-4\left(\frac{y^{4}}{a^{4}}-b^{2}+c^{2}\right)}}{2}\\ &\implies\boxed{x = c\pm\sqrt{b^2-\frac{y^{4}}{a^{4}}}} \end{align}

Answer 3

The equation is easier to work into the quadratic equation if we first get rid of radicals and avoid trying to "simplify" by using the $y^4/a^4$ rational. Expansion gives us the best view of a,b,c.

$$y=a\sqrt[^4]{b^2-(x-c)^2}\implies y^4=a^4\big(b^2-(c^2 - 2 c x + x^2)^2\big)\\ \implies 0=a^4 b^2 - a^4 c^2 + 2 a^4 c x - a^4 x^2-y^4\\ \implies (a^4)x^2 - (2 a^4 c)x + (y^4 - a^4 b^2 + a^4 c^2) = 0\\ \implies x= \frac{(2 a^4 c) \pm\sqrt{(2 a^4 c)^2-4(a^4)(y^4 - a^4 b^2 + a^4 c^2)}}{2a^4}\\ = \frac{(2 a^4 c) \pm\sqrt{4 a^8 b^2 - 4 a^4 y^4}}{2a^4} \\ \implies x = c \pm \frac{\sqrt{(a^2 b - y^2) (a^2 b + y^2)}}{a^2 } \quad a,b,c,y\in\mathbb{R}\land a\ne 0\\ x\in\mathbb{R}\iff y\le \sqrt{a^2b}$$