Lời giải chi tiết:

ĐK: \(4 - {x^2} \ge 0.\)

\( \Leftrightarrow  - 2 \le x \le 2 \Leftrightarrow x \in \left[ { - 2;2} \right].\)

+ TXĐ: \(D = \left[ { - 2;2} \right]\)

+ \(y' = 1 + \dfrac{{ - 2x}}{{2\sqrt {4 - {x^2}} }} = 1 + \dfrac{{ - x}}{{\sqrt {4 - {x^2}} }}.\)

Cho \(y' = 0 \Leftrightarrow 1 + \dfrac{{ - x}}{{\sqrt {4 - {x^2}} }} = 0.\)

 \(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\dfrac{{\sqrt {4 - {x^2}}  - x}}{{\sqrt {4 - {x^2}} }} = 0 \Leftrightarrow \,\sqrt {4 - {x^2}}  = x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,4 - {x^2} = {x^2} \Leftrightarrow \,{x^2} = 2 \Leftrightarrow \,x =  \pm \sqrt 2 \,\,\,\left( {tm} \right)\end{array}\)

Thay \(x =  - 2\)vào \(f\left( x \right)\) ta có \(f\left( { - 2} \right) =  - 2.\)

Thay \(x =  - \sqrt 2 \)vào \(f\left( x \right)\)ta có \(f\left( { - \sqrt 2 } \right) = 0.\)

Thay \(x = \sqrt 2 \)vào \(f\left( x \right)\)ta có \(f\left( {\sqrt 2 } \right) = 2\sqrt 2 .\)

Thay \(x = 2\)vào \(f\left( x \right)\)ta có \(f\left( 2 \right) = 2.\)

\( \Rightarrow \mathop {\min y}\limits_{\left[ { - 2;2} \right]}  =  - 2;\mathop {\max }\limits_{\left[ { - 2;2} \right]} \,y = 2\sqrt 2 .\)