Mathematics - Functions Question with Solution | TestHub

Given,

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

Now differentiation the above function,

$f^{\text{'}}\left(x\right)=\frac{-1}{2\sqrt{3-x}}+\frac{1}{2\sqrt{2+x}}$

Point of maxima and minima,

$f^{\text{'}}\left(x\right)=0$

$\Rightarrow \frac{\sqrt{3-x}-\sqrt{2+x}}{2\sqrt{2+x}\sqrt{3-x}}=0$

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

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

And domain is $(-2\le x\le 3)$

So at $x=\frac{1}{2}$, maxima

The maximum value of the function,

$f\left(\frac{1}{2}\right)=\sqrt{\frac{5}{2}}+\sqrt{\frac{5}{2}}=\sqrt{10}$

The minimum value of function at $x=-2 \text{or} x=3$

$f\left(-2\right)=\sqrt{5}$ and $f\left(3\right)=\sqrt{5}$

Therefore, the range of the function,

$\text{Range}\in \left[\sqrt{5},\sqrt{10}\right]$