The range of the function f(x) = 3 - x + 2 + x is

Given,

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

Now differentiation the above function,

${f}^{'} ( x ) = \dfrac{- 1}{2\sqrt{3 - x}} + \dfrac{1}{2\sqrt{2 + x}}$

Point of maxima and minima,

${f}^{'} ( x ) = 0$

$⇒ \dfrac{\sqrt{3 - x} - \sqrt{2 + x}}{2\sqrt{2 + x}\sqrt{3 - x}} = 0$

$⇒ \sqrt{2 + x} = \sqrt{3 - x}$

$⇒ x = \dfrac{1}{2}$

And domain is $( - 2 \le x \le 3 )$

So at $x = \dfrac{1}{2}$, maxima

The maximum value of the function,

$f\left(\dfrac{1}{2}\right) = \sqrt{\dfrac{5}{2}} + \sqrt{\dfrac{5}{2}} = \sqrt{10}$

The minimum value of function at $x = - 2\text{or}x = 3$

$f\left(- 2\right) = \sqrt{5}$ and $f\left(3\right) = \sqrt{5}$

Therefore, the range of the function,

$\text{Range} \in \left[\sqrt{5}, \sqrt{10}\right]$