Here is an interesting question that asks whether there is any function that satisfies $f^{-1}=f'\;$ i.e. its derivative is equal to the inverse function. In addition to the answers provided there, it can be shown that this can be reduced to solving the delay differential equation $$g'(x) =\frac{1}{g(x-1)}g'(x+1)$$ I am not aware of any practical use of this kind of equations, but it has a nice form nonetheless! and the functional can be defined as $f(g(x))=g(x+1)$.

Anyway, here's a fairly simple solution which is a bijection on $\mathbb R$ and, lo and behold, is entirely based on the golden ratio: $$f(x)=\cases{\begin{array}{rl} \phi (x/\phi)^{\phi } & x\geq 0 \\ \psi (x/\psi)^{\psi } & x<0 \text{ where } \psi =-\phi ^{-1} \\ \end{array}}$$

Here is an interesting question that asks whether there is any function that satisfies $f^{-1}=f'\;$ i.e. its derivative is equal to the inverse function. In addition to the answers provided there, it can be shown that this can be reduced to solving the delay differential equation $$g'(x) =\frac{1}{g(x-1)}g'(x+1).$$ I am not aware of any practical use of this kind of equations, but it has a nice form nonetheless! And the functional can be defined as $f(g(x))=g(x+1)$.

Anyway, here's a fairly simple solution which is a bijection on $\mathbb R$ and, lo and behold, is entirely based on the golden ratio: $$f(x)=\begin{cases} \phi (x/\phi)^{\phi } & x\geq 0 \\ \psi (x/\psi)^{\psi } & x<0 \text{ where } \psi =-\phi ^{-1}. \\ \end{cases}$$

Here is an interesting question that asks whether there is any function that satisfies $f^{-1}=f'\;$ i.e. its derivative is equal to the inverse function. In addition to the answers provided there, it can be shown that this can be reduced to solving the delay differential equation $$g'(x) =\frac{1}{g(x-1)}g'(x+1)$$ I am not aware of any practical use of this kind of equations, but it has a nice form nonetheless! and the functional can be defined as $f(g(x))=g(x+1)$.

Anyway, here's a fairly simple solution which is a bijection on $\mathbb R$ and, lo and behold, is entirely based on the golden ratio: $$f(x)=\cases{\begin{array}{rl} \phi x^{\phi } & x\geq 0 \\ \psi x^{\psi } & x<0 \text{ where } \psi =-\phi ^{-1} \\ \end{array}}$$

Here is an interesting question that asks whether there is any function that satisfies $f^{-1}=f'\;$ i.e. its derivative is equal to the inverse function. In addition to the answers provided there, it can be shown that this can be reduced to solving the delay differential equation $$g'(x) =\frac{1}{g(x-1)}g'(x+1)$$ I am not aware of any practical use of this kind of equations, but it has a nice form nonetheless! and the functional can be defined as $f(g(x))=g(x+1)$.

Anyway, here's a fairly simple solution which is a bijection on $\mathbb R$ and, lo and behold, is entirely based on the golden ratio: $$f(x)=\cases{\begin{array}{rl} \phi (x/\phi)^{\phi } & x\geq 0 \\ \psi (x/\psi)^{\psi } & x<0 \text{ where } \psi =-\phi ^{-1} \\ \end{array}}$$