The range of the function f(x) = sqrt{3-x} + | vedclass

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(A) Let $y = \sqrt{3-x} + \sqrt{2+x}$. The domain is determined by $3-x \ge 0$ and $2+x \ge 0$,which gives $-2 \le x \le 3$.Squaring both sides,we get

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$[\sqrt{5}, \sqrt{10}]$

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(A) Let $y = \sqrt{3-x} + \sqrt{2+x}$. The domain is determined by $3-x \ge 0$ and $2+x \ge 0$,which gives $-2 \le x \le 3$.Squaring both sides,we get $y^2 = (3-x) + (2+x) + 2\sqrt{(3-x)(2+x)} = 5 + 2\sqrt{6+x-x^2}$.To find the range,we analyze the expression $g(x) = 6+x-x^2$. This is a downward-opening parabola with vertex at $x = -\frac{b}{2a} = -\frac{1}{2(-1)} = \frac{1}{2}$.The maximum value of $g(x)$ is $g(1/2) = 6 + 1/2 - 1/4 = 6 + 1/4 = 25/4$.The minimum value of $g(x)$ occurs at the endpoints of the domain,$x = -2$ or $x = 3$. $g(-2) = 6-2-4 = 0$ and $g(3) = 6+3-9 = 0$.Thus,$0 \le g(x) \le 25/4$.Substituting this into $y^2 = 5 + 2\sqrt{g(x)}$,we get $5 + 2\sqrt{0} \le y^2 \le 5 + 2\sqrt{25/4}$.$5 \le y^2 \le 5 + 2(5/2) = 10$.Since $y \ge 0$,the range is $[\sqrt{5}, \sqrt{10}]$.

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

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