The range of the function $$f(x) = \sqrt{3-x} + \sqrt{2+x}$$ is

$$f(x) = \sqrt{3-x} + \sqrt{2+x}$$. Domain: $$-2 \leq x \leq 3$$.

Square: $$f(x)^2 = (3-x) + (2+x) + 2\sqrt{(3-x)(2+x)} = 5 + 2\sqrt{(3-x)(2+x)}$$.

Let $$g(x) = (3-x)(2+x) = -x^2 + x + 6$$. This is a downward parabola with max at $$x = 1/2$$.

$$g(1/2) = (5/2)(5/2) = 25/4$$.

$$g(-2) = 5 \cdot 0 = 0$$ and $$g(3) = 0 \cdot 5 = 0$$.

So $$g(x) \in [0, 25/4]$$, and $$\sqrt{g(x)} \in [0, 5/2]$$.

$$f(x)^2 \in [5, 5 + 5] = [5, 10]$$.

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