Tập xác định D=[-2;2], \(f'(x)=-\frac{x}{\sqrt{4-x^2}}+1\) \(f'(x)=0\Leftrightarrow -\frac{x}{\sqrt{4-x^2}}+1=0\Leftrightarrow \sqrt{4-x^2}=x\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ 4-x^2=x^2 \end{matrix}\right.\) \(\Leftrightarrow x=\sqrt{2}\) Ta có \(f(\sqrt{2})=2\sqrt{2};f(2)=2;f(-2)=-2,f(3)=7\) Vậy \(\underset{[-2;2]}{Max}=2\sqrt{2}\) khi \(x=\sqrt{2}\) \(\underset{[-2;2]}{Min}=-2\) khi x = -2