Someone gave me a random maths problem to solve:
Given that $h \left ( \dfrac{x}{x^2+h(x)} \right )=1$, what is $h(x)$
The restrictions given were:
- $h(x) \neq constant$
- $\exists \frac{dh}{dx}$
- $\exists h^{-1}(x)$
- $\exists \frac{dh^{-1}}{dx}$
However, I am not sure how to go about solving this. I started off with:
$$\begin{align*} \frac{x}{x^2+h(x)} &= h^{-1}(1)\\\\ \frac{x^2+h(x)}{x} &= \frac{1}{h^{-1}(1)}\\\\ x^2+h(x) &= \frac{x}{h^{-1}(1)}\\\\ h(x) &= \frac{x}{h^{-1}(1)} - x^2 \end{align*}$$
which doesn't feel like it gets me any closer to something that makes sense. I don't even know what keywords to do some Google searching with in order to educate myself on the kind of manipulations that can be done to solve this kind of question (searching for "functions that call themselves" or "recursive function" generate tons of what look like entirely unrelated results).
What do I need to know in order to be able to tackle this kind of problem?
Assuming $h : \mathbb{R} \to \mathbb{R}$, there is trivially no such $h$. Since $h^{-1}$ exists, you correctly deduced that $$ \frac{x}{x^2 + h(x)} = h^{-1}(1) \;\; \forall \;x. $$ Plugging in $x = 0$, $h^{-1}(1) = 0$. But plugging in $x \ne 0$, a nonzero number over a real number then equals zero, contradiction.
EDIT: In fact, the exact same argument shows there is no $h$ defined for $x = 0$ and at least one other complex number.