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

Question

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

Accepted Answer

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

$$y' = {{ - 1} \over {2\sqrt 3  - x}} + {1 \over {2\sqrt {x + 2} }} = 0$$

$$ \Rightarrow \sqrt x  + 2 = \sqrt 3  - x$$

$$ \Rightarrow x = {1 \over 2}$$

$$y\left( {{1 \over 2}} \right) = \sqrt {{5 \over 2}}  + \sqrt {{5 \over 2}}  = \sqrt {10} $$

$${y_{\min }}$$ at $x =  - 2$ or $x = 3$, i.e., $$\sqrt 5 $$

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

Solutions · 1