Lời giải chi tiết:

ĐK: \(1 - {x^2} \ge 0 \Leftrightarrow  - 1 \le x \le 1 \Leftrightarrow \,x \in \left[ { - 1;1} \right]\)

+ TXĐ: \(D = \left[ { - 1;1} \right]\)

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

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

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow 1 - {x^2} = \,{x^2} \Leftrightarrow \,{x^2} = \dfrac{1}{2}\, \Leftrightarrow \,x =  \pm \sqrt {\dfrac{1}{2}} \,\,\,\,\left( {tm} \right)\)

Thay \(x =  - 1\)vào \(f\left( x \right)\) ta có \(f\left( { - 1} \right) = 0.\)

Thay \(x = \left( { - \sqrt {\dfrac{1}{2}} } \right)\) vào \(f\left( x \right)\)ta có \(f\left( { - \sqrt {\dfrac{1}{2}} } \right) = \dfrac{{ - 1}}{2}.\)

Thay \(x = \sqrt {\dfrac{1}{2}} \)vào \(f\left( x \right)\)ta có \(f\left( {\sqrt {\dfrac{1}{2}} } \right) = \dfrac{1}{2}.\)

Thay \(x = 1\)vào \(f\left( x \right)\)ta có \(f\left( 1 \right) = 0.\)

\( \Rightarrow \mathop {\min y}\limits_{\left[ { - 1;1} \right]}  =  - \dfrac{1}{2};\mathop {\max y}\limits_{\left[ { - 1;1} \right]}  = \dfrac{1}{2}.\)