Answer
The maximum is $f(4) \approx0.47213$ and the minimum is $f(\sqrt{3}/2 ) \approx -0.589980$
Work Step by Step
Given $$y=\sqrt{x+x^{2}}-2 \sqrt{x}, \quad[0,4]$$
Since
\begin{align*}
\frac{dy}{dx}&=\frac{1+2x}{2\sqrt{x+x^2}}-\frac{1}{\sqrt{x}}\\
&=\frac{1+2 x-2 \sqrt{1+x}}{2 \sqrt{x} \sqrt{1+x}}
\end{align*}
Find the critical points
\begin{align*}
f'(x)&=0\\
\frac{1+2 x-2 \sqrt{1+x}}{2 \sqrt{x} \sqrt{1+x}}&=0\\
1+2 x-2 \sqrt{1+x}&=0\\
4x^2-3&=0
\end{align*}
Then $x=\pm\sqrt{3}/2$. Because $-\sqrt{3}/2\notin [0,4]$, then $f(x)$ has a critical point $\sqrt{3}/2$, since
\begin{align*}
f(0)&=0\\
f(\sqrt{3}/2 )&\approx -0.589980\\
f(4)&\approx0.47213
\end{align*}
Then the maximum is $f(4) \approx0.47213$ and the minimum is $f(\sqrt{3}/2 ) \approx -0.589980$
