The range of the function $f(x)=\sqrt{3-x}+\sqrt{2+x}$ is :
Solution
<p>$f(x) = \sqrt {3 - x} + \sqrt {x + 2}$</p>
<p>$y' = {{ - 1} \over {2\sqrt 3 - x}} + {1 \over {2\sqrt {x + 2} }} = 0$</p>
<p>$\Rightarrow \sqrt x + 2 = \sqrt 3 - x$</p>
<p>$\Rightarrow x = {1 \over 2}$</p>
<p>$$y\left( {{1 \over 2}} \right) = \sqrt {{5 \over 2}} + \sqrt {{5 \over 2}} = \sqrt {10} $$</p>
<p>${y_{\min }}$ at $x = - 2$ or $x = 3$, i.e., $\sqrt 5$</p>
<p>$\therefore$ $f(x) \in [\sqrt 5 ,\sqrt {10} ]$</p>
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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