$$f(x) = \sqrt {3 - x} + \sqrt {x + 2} $$

$$y' = {{ - 1} \over {2\sqrt 3 - x}} + {1 \over {2\sqrt {x + 2} }} = 0$$

$$ \Rightarrow \sqrt x + 2 = \sqrt 3 - x$$

$$ \Rightarrow x = {1 \over 2}$$

$$y\left( {{1 \over 2}} \right) = \sqrt {{5 \over 2}} + \sqrt {{5 \over 2}} = \sqrt {10} $$

$${y_{\min }}$$ at $$x = - 2$$ or $$x = 3$$, i.e., $$\sqrt 5 $$

$$\therefore$$ $$f(x) \in [\sqrt 5 ,\sqrt {10} ]$$