5
$\begingroup$

This question already has an answer here:

Find an example of an inverse function f(x) such that its derivative is the same as its inverse.

I tried many different functions but non of them worked.

$\endgroup$

marked as duplicate by Amzoti, Asaf Karagila, Julian Kuelshammer, Paul, Stahl May 14 '13 at 5:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

We find a function defined for all positive $x$, with the right properties.

There may be no general theory, so let's fool around and look for a function of the shape $f(x)= cx^a$, for constants $a$ and $c$.

Then the derivative of $f$ is given by $f'(x)=acx^{a-1}$. The inverse function is given by $f^{-1}(x)=\frac{x^{1/a}}{c^{1/a}}$. So we need $$acx^{a-1}=\frac{x^{1/a}}{c^{1/a}}.$$ For the the identity above to hold, we want $a-1=\frac{1}{a}$. Rewrite as $a^2-a-1=0$, which has $a=\frac{1+\sqrt{5}}{2}$ as a root. Nice number!

We also want $ca=\frac{1}{c^{1/a}}$. There is such a $c$. It is given by $$c=\left(\frac{\sqrt{5}-1}{2}\right)^{1/\sqrt{5}}.$$

$\endgroup$
  • $\begingroup$ Very very nice! $\endgroup$ – please delete me May 14 '13 at 4:09
3
$\begingroup$

If I understood your question correctly and assuming that you are asking about functions over $R,$ there is no such function. Indeed, if $f'(x)=g(x)$ and $f(g(x))=x=g(f(x))$ then $f'(f(x))=x.$ But $f$ has to be a bijection so it is monotone and derivative cannot change the sign.

$\endgroup$
  • $\begingroup$ But this was the way the question was worded in the maths competition. $\endgroup$ – please delete me May 14 '13 at 3:53

Not the answer you're looking for? Browse other questions tagged or ask your own question.