$\begin{array}{l} f\left( x \right) = \left( {x - 1} \right)\sqrt {3 + 2x - {x^2}} \\ TXD:\,\,\,D = \left[ { - 1;\,\,\,3} \right].\\ \Rightarrow f'\left( x \right) = \sqrt {3 + 2x - {x^2}} + \frac{{\left( { - 2x + 2} \right)\left( {x - 1} \right)}}{{2\sqrt {3 + 2x - {x^2}} }}\\ = \sqrt {3 + 2x - {x^2}} - \frac{{{{\left( {x - 1} \right)}^2}}}{{\sqrt {3 + 2x - {x^2}} }}\\ = \frac{{3 + 2x - {x^2} - {x^2} + 2x - 1}}{{\sqrt {3 + 2x - {x^2}} }} = \frac{{ - 2{x^2} + 4x + 2}}{{\sqrt {3 + 2x - {x^2}} }}\\ \Rightarrow f'\left( x \right) = 0\\ \Leftrightarrow - 2{x^2} + 4x + 2 = 0 \Leftrightarrow \left[ \begin{array}{l} x = 1 - \sqrt 2 \\ x = 1 + \sqrt 2 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} f\left( { - 1} \right) = 0\\ f\left( {1 - \sqrt 2 } \right) = - 2\\ f\left( {1 + \sqrt 2 } \right) = 2\sqrt 2 \\ f\left( 3 \right) = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \mathop {Max}\limits_{\left[ { - 1;\,\,3} \right]} f\left( x \right) = f\left( {1 + \sqrt 2 } \right) = 2\sqrt 2 \\ \mathop {Min}\limits_{\left[ { - 1;\,3} \right]} f\left( x \right) = f\left( {1 - \sqrt 2 } \right) = - 2 \end{array} \right.. \end{array}$