Range for given map

Exercise:

Determine the values of $\alpha $ such that $\sqrt{3-\cos \frac{1}{x}}=3-\alpha $

Solution: we know that $\left| \cos u\left( x \right) \right|\le 1\Leftrightarrow -1\le -\cos u\left( x \right)\le 1$

$\Rightarrow 3-1\le 3-\cos u\left( x \right)\le 3+1\Leftrightarrow 2\le 3-\cos \frac{1}{x}\le 4\Leftrightarrow \sqrt{2}\le \sqrt{3-\cos \frac{1}{x}}\le 2$

But $\sqrt{3-\cos \frac{1}{x}}=3-\alpha \Leftrightarrow \sqrt{2}\le 3-\alpha \le 2\Leftrightarrow \sqrt{2}-3\le -\alpha \le 2-3\Leftrightarrow \sqrt{2}-3\le -\alpha \le -1$

$\Rightarrow 1\le \alpha \le 3-\sqrt{2}$ hence $\alpha \in \left[ 1\,\,,\,\,\,3-\sqrt{2} \right]$