This question mostly eluded me during the exam itself.
Problem: Suppose that $f: [-1, 1] \rightarrow \mathbb{R}$ continuously, and that
$$\begin{align} \text{(i)}\qquad &f(x) = \frac{2 - x^2}{2} f\left(\frac{x^2}{2-x^2}\right)\\ \text{(ii)}\qquad &f(0) = 1\\[6pt] \text{(iii)}\qquad &\lim_{x\to 1^-}\frac{f(x)}{\sqrt{1-x}}\ \text{exists} \end{align}$$
Determine the closed form of $f$, and prove that it is unique. ---
All that I managed to do was - using the eventual heuristic of "powers of $1 - x^2$ will cancel neatly in (iii) and work in (ii), so let's see if they work in (i)" - discover that
$$ f(x) = \sqrt{1 - x^2}$$
Now, this $f$ is an involution, so it suffices to prove that if $g$ is a function satisfying (i) to (iii), then $g(f(x)) = x$. This, however, was more than I could do.
How do you prove uniqueness? (I ran across one solution I did not understand; please, be gentle.)