Question

The maximum value of the function $f\left(x\right)=2\sqrt{x-2}+\sqrt{4-x}$ is $\sqrt{K},$ then $K$ is

Answer 1

Clearly, domain of the function is $[2,4].$ Now, $f^{\prime }\left(x\right)=\frac{1}{\sqrt{x-2}}-\frac{1}{2\sqrt{4-x}}$ $f^{\prime }\left(x\right)=0$ $\Rightarrow \sqrt{x-2}=2\sqrt{4-x}$ $\Rightarrow x-2=16-4x$
$x=\frac{18}{5}$
Now, $f\left(2\right)=\sqrt{2},f\left(\frac{18}{5}\right)=2\sqrt{\frac{18}{5}-2}+\sqrt{4-\frac{18}{5}}=\sqrt{10},$
$f\left(4\right)=2\sqrt{2}$
By above obtained values, we can say that $x=\left(\frac{18}{5}\right)$is the point of global maxima.
Hence, range of the function is $[\sqrt{2},\sqrt{10}].$