find the range of √(2-x) + √(1+x)

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$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\sqrt{2-\mathrm{x}}+\sqrt{1+\mathrm{x}} \\ \mathrm{Quantity} \mathrm{inside} \mathrm{root} \mathrm{must} \mathrm{be} \mathrm{positive} \\ 2-\mathrm{x}\ge 0 \mathrm{and} 1+\mathrm{x}\ge 0 \\ \Rightarrow \mathrm{x}\le 2 \mathrm{and} \mathrm{x}\ge -1 \\ \mathrm{Take} \mathrm{common} \mathrm{values} \\ \mathrm{Hence} \mathrm{domain} \mathrm{is} \mathrm{x}\in \left[-1,2\right] \\ \mathrm{f}\left(-1\right)=\sqrt{2+1}+0=\sqrt{3} \\ \mathrm{f}\left(2\right)=0+\sqrt{1+2}=\sqrt{3} \\ \mathrm{Differentiate} \mathrm{f}\left(\mathrm{x}\right) \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=-\frac{1}{2\sqrt{2-\mathrm{x}}}+\frac{1}{2\sqrt{1+\mathrm{x}}} \\ \mathrm{For} \mathrm{maxima}/\mathrm{mnima}, \mathrm{f}\text{'}\left(\mathrm{x}\right)=0 \\ \Rightarrow \frac{1}{2\sqrt{1+\mathrm{x}}}=\frac{1}{2\sqrt{2-\mathrm{x}}} \\ \Rightarrow 2-\mathrm{x}=1+\mathrm{x} \\ \Rightarrow 2\mathrm{x}=3 \\ \Rightarrow \mathrm{x}=\frac{3}{2} \\ \mathrm{Hence} \\ \mathrm{f}\left(\frac{3}{2}\right)=\sqrt{2-\frac{3}{2}}+\sqrt{1+\frac{3}{2}}=\frac{1}{\sqrt{2}}+\frac{5}{\sqrt{2}}=\frac{6}{\sqrt{2}}=3\sqrt{2} \\ \mathbf{Hence}\mathbf{ }\mathbf{domain}\mathbf{ }\mathbf{is}\mathbf{ }\left[\sqrt{\mathbf{3}}\mathbf{,}\mathbf{3}\sqrt{\mathbf{2}}\right] \end{gathered}$$

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