Let f(x)=3 sqrt{x-2}+sqrt{4-x} be a real-valu | vedclass

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(B) Given the function $f(x)=3 \sqrt{x-2}+\sqrt{4-x}$.The domain of the function is determined by $x-2 \geq 0$ and $4-x \geq 0$,which implies $x \in [

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$42$

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(B) Given the function $f(x)=3 \sqrt{x-2}+\sqrt{4-x}$.The domain of the function is determined by $x-2 \geq 0$ and $4-x \geq 0$,which implies $x \in [2, 4]$.To find the range,we use the Cauchy-Schwarz inequality or trigonometric substitution. Let $u = \sqrt{x-2}$ and $v = \sqrt{4-x}$. Then $u^2 + v^2 = (x-2) + (4-x) = 2$.The function is $f(x) = 3u + v$. By the Cauchy-Schwarz inequality,$(3u + v)^2 \leq (3^2 + 1^2)(u^2 + v^2) = (9+1)(2) = 20$.Thus,$f(x)^2 \leq 20$,so $f(x) \leq \sqrt{20} = \beta$.At the boundaries of the domain:If $x=2$,$f(2) = 3(0) + \sqrt{2} = \sqrt{2}$.If $x=4$,$f(4) = 3\sqrt{2} + 0 = 3\sqrt{2} = \sqrt{18}$.Since $\sqrt{2} Therefore,$\alpha^2 + 2\beta^2 = (\sqrt{2})^2 + 2(\sqrt{20})^2 = 2 + 2(20) = 2 + 40 = 42$.

Let $f(x)=3 \sqrt{x-2}+\sqrt{4-x}$ be a real-valued function. If $\alpha$ and $\beta$ are respectively the minimum and the maximum values of $f$,then $\alpha^2+2 \beta^2$ is equal to

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