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Q.
The range of the function $f(x)=2 \sqrt{x -2} + \sqrt{4 - x}$ is
Application of Derivatives
Solution:
Clearly, domain of the function is $[2,4]$. Now,
$f'(x)=\frac{1}{\sqrt{x-2}}-\frac{1}{2 \sqrt{4-x}}$
$f'(x)=0 $ or $\sqrt{x-2}=2 \sqrt{4-x}$
or $x-2=16-4 x$ or
$x=\frac{18}{5}$
Now, $f(2)=\sqrt{2}, f\left(\frac{18}{5}\right)$
$=2 \sqrt{\frac{18}{5}-2}+\sqrt{4-\frac{18}{5}}=\sqrt{10}$
$f(4)=2 \sqrt{2}$
Hence, range of the function is $[\sqrt{2}, \sqrt{10}]$.
Also, here $x=(18 / 5)$ is the point of global maxima.