## Calculus (3rd Edition)

The maximum is $f(2)=4\sqrt{2}$ and the minimum is $f\left( 0\right) =2\sqrt{6}$
Given $$y=(2+x) \sqrt{2+(2-x)^{2}}, \quad[0,2]$$ Since \begin{align*} \frac{dy}{dx}&=\frac{d}{dx}\left(2+x\right)\sqrt{2+\left(2-x\right)^2}+\frac{d}{dx}\left(\sqrt{2+\left(2-x\right)^2}\right)\left(2+x\right)\\ &=\frac{2x^2-4x+2}{\sqrt{x^2-4x+6}} \end{align*} Find critical points \begin{align*} f'(x)&=0\\ \frac{2x^2-4x+2}{\sqrt{x^2-4x+6}}&=0\\ 2x^2-4x+2&=0\\ 2(x-1)^2&=0 \end{align*} Then $x=1$. Because $1\in [0,2]$, then $f(x)$ has a critical point at $x= 1$, since \begin{align*} f(0)&=2\sqrt{6}\\ f(1)&= 3\sqrt{3}\\ f(2)&=4\sqrt{2} \end{align*} Then the maximum is $f(2)=4\sqrt{2}$ and the minimum is $f\left( 0\right) =2\sqrt{6}$