# Inverse of fourth root function

Mathematics Asked by Sean Xie on February 28, 2021

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

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!

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.

Correct answer by PierreCarre on February 28, 2021

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.

$$y=asqrt[^4]{b^2-(x-c)^2}implies y^4=a^4big(b^2-(c^2 - 2 c x + x^2)^2big)\ implies 0=a^4 b^2 - a^4 c^2 + 2 a^4 c x - a^4 x^2-y^4\ implies (a^4)x^2 - (2 a^4 c)x + (y^4 - a^4 b^2 + a^4 c^2) = 0\ implies x= frac{(2 a^4 c) pmsqrt{(2 a^4 c)^2-4(a^4)(y^4 - a^4 b^2 + a^4 c^2)}}{2a^4}\ = frac{(2 a^4 c) pmsqrt{4 a^8 b^2 - 4 a^4 y^4}}{2a^4} \ implies x = c pm frac{sqrt{(a^2 b - y^2) (a^2 b + y^2)}}{a^2 } quad a,b,c,yinmathbb{R}land ane 0\ xinmathbb{R}iff yle sqrt{a^2b}$$

Answered by poetasis on February 28, 2021

begin{align} y=asqrt[^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^2-4left(frac{y^{4}}{a^{4}}-b^{2}+c^{2}right)}}{2}\ &impliesboxed{x = cpmsqrt{b^2-frac{y^{4}}{a^{4}}}} end{align}

Answered by DMcMor on February 28, 2021