Vietnamese Mathematical Olympiad, 2003

We have:
$$f(\cot x) = \sin 2x + \cos 2x \quad \forall x \in (0; \pi)$$
$$\Leftrightarrow f(\cot x) = \frac{2 \cot x}{\cot^2 x + 1} + \frac{\cot^2 x - 1}{\cot^2 x + 1} = \frac{\cot^2 x + 2 \cot x - 1}{\cot^2 x + 1} \quad \forall x \in (0; \pi).$$
Therefore, remarking that for every $t \in \mathbb{R}$ there exists $x \in (0; \pi)$ such that $\cot x = t$, we get:
$$f(t) = \frac{t^2 + 2t - 1}{t^2 + 1} \quad \forall t \in \mathbb{R}.$$
It implies
$$g(x) = f(x) \cdot f(1-x) = \frac{x^2(1-x)^2 + 8x(1-x) - 2}{x^2(1-x)^2 - 2x(1-x) + 2} \quad \forall x \in \mathbb{R}. \quad (*)$$
Put $u = x(1-x)$. It is easy to see that when $x$ runs through $[-1; 1]$, $u$ runs through $[-2; 1/4]$. So, from (*) we have:
$$\begin{array}{l} \min g(x) = \min h(u) \quad \text{and} \quad \max g(x) = \max h(u), \\ -1 \le x \le 1 \qquad -2 \le u \le 1/4 \qquad -1 \le x \le 1 \qquad -2 \le u \le 1/4 \end{array}$$
where $h(u) = \frac{u^2 + 8u - 2}{u^2 - 2u + 2}$.
By studying the sign of $h'(u) = \frac{2(-5u^2 + 4u + 6)}{(u^2 - 2u + 2)^2}$ on $[-2; 1/4]$, we get:
$$\min_{-2 \le u \le 1/4} h(u) = h((2 - \sqrt{34})/5) = 4 - \sqrt{34}, \text{ and}$$
$$\max_{-2 \le u \le 1/4} h(u) = \max\{h(-2), h(1/4)\} = \max\{-7/5, 1/25\} = 1/25.$$
So, on $[-1; 1]$, $\min g(x) = 4 - \sqrt{34}$, $\max g(x) = 1/25$.