Inverse of fourth root function

Question

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!

PierreCarre · Accepted Answer

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.

poetasis · Answer

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

DMcMor · Answer

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}

Inverse of fourth root function

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