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Let's say $f:D\to R$ is an injective function on some domain where it is also differentiable. For a real function, i.e. $D\subset\mathbb R, R\subset\mathbb R$, is it possible that $f'(x)\equiv f^{-1}(x)$?
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?
It is possible! Here is an example on the domain $D=[0,\infty)$: $$ f(x) = \bigg(\frac{\sqrt{5}-1}{2}\bigg)^{(\sqrt5-1)/2} x^{(\sqrt5+1)/2}. $$ I found this by supposing that $f(x)$ had the form $ax^b$, setting the derivative equal to the inverse function, and solving for $a$ and $b$.
On $(0, \infty)$, take $f(x) = a x^p$ where $p = (\sqrt{5}+1)/2$ (so that $p(p-1) = 1$) and $a = p^{-1/p}$.
There have already been examples with $f: D \to \mathbb R$, but note that it is not possible with $f:\mathbb R \to \mathbb R$. A simple argument is that for a function $f$ to be injective, necessarily $f'(x) \geq 0$ or $f'(x) \leq 0$ for all $x$. Thus we can see that for there to be equality between $f'(x)$ and $f^{-1}(x)$, then we must have $f^{-1}(x) \geq 0$ or $f^{-1}(x) \leq 0$ for all $x$.
But this can't happen, because any function defined on $f: \mathbb R \to \mathbb R$ must have its inverse go from positive to negative for some $x$. To confirm this, just look at the fact that the inverse of any horizontal line must cross the x-axis by flipping over the line $y=x$, and then add curves to that line to find that nothing has changed, and it still must cross the x-axis.
\to (usually synonym to \rightarrow), which renders as $\to$.
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\mapsto is used for denoting a single value mapping. For example, $f: C\to X$ expresses we mean a function $f$ going from a set $C$ to a set $X$, while $q: x\mapsto \sin (x/2)$ expresses we mean a function $q$ which assigns each $x$ argument a value of a sine of a half of $x$. See Wikipedia article 'List of mathematical symbols', section 'Symbols that point left or right'
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Since someone mentioned whether the answer given by Robert Israel / Greg Martin might be unique, I thought it is worth noting that the function $$f(x)=-\frac{1}{\phi^\phi}(-x)^{-\frac{1}{\phi}},\quad x<0$$ where $\phi$ is the golden ratio, has the same property on $D=(-\infty,0)$, i.e. $f'(x)\equiv f^{-1}(x)$.
Edit- So if we define: $$f:\mathbb R\to\mathbb R\\x\mapsto a(x/a)^a$$ where $a=\frac{1+\sqrt{5}\text{ sign}(x)}{2}$, we would have a bijection on $\mathbb R$ with that nice property (yay!...).
