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Is there any function, $f(x)\neq x$, for which $f(f'(x))=f'(f(x))$?


marked as duplicate by YuiTo Cheng, Ak19, cmk, Shogun, metamorphy Jun 23 at 18:53

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For the real-valued function $f(x)=e^x$, $f'(x)=e^x$.

  • $\begingroup$ Thanks. $f(x)=e^x$ works. What if the condition was $f(x) \neq f'(x) \neq x$? $\endgroup$ – Hussain-Alqatari Jun 19 at 8:21
  • 1
    $\begingroup$ There are still many. $\endgroup$ – Ivan Neretin Jun 19 at 8:24

Why, many. $f(x)={x^2\over2}$ will do.

If you want to exclude the "trick" answers where $f'(x)=x$ or $f(x)=f'(x)$, go with $f(x)={x^3\over9}$. Other examples are still plenty, I believe.

Upd. The other examples seem less numerous than I initially believed, but anyway, $f(x)=\dfrac{x^n}{n^{n-1}}$ for any constant $n$ is good.


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