Inverse of $y=\frac{1}{27}(x^5+2x^3)$?

Question

Is it possible to find the inverse of a function such as $y=\frac{1}{27}(x^5+2x^3)$? This question was posed to me by an AP Calculus AB student. The function seems to be $1$-to-$1$ and have an inverse. However, I don't know a way to solve it. Symbolab doesn't seem to know either. The student was recalling the question from memory so I'm not $100\%$ sure this was the exact question. He also mentioned that a piece of information, $a=7$, might be a part of it. He was also under the impression that the question had to do with derivatives. Thoughts?

Answer 1

I apologize for the vague question, but I think I've figured out what's going on here. Looking at the graph of any function with it's inverse, it's clear that if the slope of the tangent line at point $(a,b)$ is $m$, then the slope of the tangent line of the inverse function at point $(b,a)$ is $1/m$.

In this case, the derivative function would be $y'=\frac{1}{27}(5x^4+6x^2)$ and $y'(7)$ would be $\frac{12299}{27}$. The $b$ value would be $\frac{1}{27}((7)^5+2(7)^3)=\frac{5138}{9}$ and therefore $(y^{-1})'(\frac{5138}{9})=\frac{1}{\frac{12299}{27}}=\frac{27}{12299}$.

Answer 2

While the function is indeed one-to-one, it is impossible in general to find an analytic expression for $x$ even at $y=1$ – the polynomial that has to be solved in that case, $x^5+2x^3-27$, has Galois group $S_5$.

Nevertheless, since the function is monotone and odd, Newton's method works very well to obtain a numerical result.