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I have tried extensively to find the inverse of $f(x)=x^{0.2}+x^{0.5}$, but to no avail. I asked my algebra professor, and she didn't know either. I put the question into Wolfram Alpha, but it said "no results found in terms of standard mathematical functions". What is the inverse? Is it even possible?

Daniel Williams
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  • My first guess would be to try a substitution $y = \sqrt[10]{x}$ but that may be a dead end. Then the equation would become $f(y) = y^2 + y^5$ (not taking at all into account any sign issues with the radicals). – John Sep 21 '16 at 19:20
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    The function is defined for nonnegative numbers. Its derivate is $0.2x^{-0.8}+0.5x^{-0.5}$, which is positive for all $x>0$, so it has an inverse, but I doubt it can be expressed with elementary functions. – Peter Sep 21 '16 at 19:22
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    To expand on what @Peter said, there are lots of functions that we can describe, and prove that they exist, but we cannot write an explicit expression for then. The famous example is "the function $f$ such that $f(0)=0$ and $f'(x)=e^{-x^2}$", but there are many more. Your inverse function is probably one of them. – Arthur Sep 21 '16 at 19:27
  • I think it can be rewritten as $f(x) = \sum_n [(0.2^n+0.5^n)ln(x)^n)/n!]$ so the coefficients are a bit intertwined... – Emil Sep 21 '16 at 19:48
  • $(0.2^n+0.5^n)^{1/n}$ seem to be between 0.5 and 0.7 at least which could give you a starting guess if you'll satisfy with a non-closed formula :) – Emil Sep 21 '16 at 20:01

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