Lời giải chi tiết:

+ TXĐ: \(D = \left[ {0;2} \right]\)

+\(y' = f'\left( x \right) = \dfrac{{{{\left( {2{x^2} - 3x + 3} \right)}^\prime }.\left( {x + 1} \right) - \left( {2{x^2} - 3x + 3} \right).{{\left( {x + 1} \right)}^\prime }}}{{{{\left( {x + 1} \right)}^2}}}\)             \(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\left( {4x - 3} \right).\left( {x + 1} \right) - \left( {2{x^2} - 3x + 3} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{4{x^2} + 4x - 3x - 3 - 2{x^2} + 3x - 3}}{{{{\left( {x + 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2{x^2} + 4x - 6}}{{{{\left( {x + 1} \right)}^2}}}.\end{array}\)

Cho \(y' = 0\)\( \Leftrightarrow \dfrac{{2{x^2} + 4x - 6}}{{{{\left( {x + 1} \right)}^2}}} = 0 \Leftrightarrow 2{x^2} + 4x - 6 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 1\,\,\,\,\,\,\,\left( {tm} \right)\\x =  - 3\,\,\,\left( {Loai} \right)\end{array} \right.\)

Thay \(x = 0\)vào \(f\left( x \right)\) ta có :\(f\left( 0 \right) = 3.\)

Thay \(x = 2\)vào \(f\left( x \right)\) ta có :\(f\left( 2 \right) = \dfrac{5}{3}.\)

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

\(\max f\left( x \right) = 3;\,\,\min f\left( x \right) = 1\).