Can there be an injective function whose derivative is equivalent to its inverse function?

Let’s say f:DR is an injective function on some domain where it is also differentiable. For a real function, i.e. DR,RR, is it possible that f(x)f1(x)?

Intuitively speaking, I suspect that this is not possible, but I can’t provide a reasonable proof since I know very little nothing about functional analysis. Can anyone provide a (counter)example or prove that such function does not exist?

Answer

It is possible! Here is an example on the domain D=[0,):
f(x)=(512)(51)/2x(5+1)/2.
I found this by supposing that f(x) had the form axb, setting the derivative equal to the inverse function, and solving for a and b.

graph of $f(x)$ and $f'(x)$

Attribution
Source : Link , Question Author : polfosol , Answer Author : Greg Martin

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