Lời giải chi tiết:

+ TXĐ: \(D = \left[ {0;2} \right]\)

+ \(y' = f'\left( x \right) = \dfrac{{{{\left( {4{x^2} + 7x + 7} \right)}^\prime }.\left( {x + 2} \right) - {{\left( {x + 2} \right)}^\prime }.\left( {4{x^2} + 7x + 7} \right)}}{{{{\left( {x + 2} \right)}^2}}}\)

   \(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\left( {8x + 7} \right).\left( {x + 2} \right) - \left( {4{x^2} + 7x + 7} \right)}}{{{{\left( {x + 2} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{8{x^2} + 16x + 7x + 14 - 4{x^2} - 7x - 7}}{{{{\left( {x + 2} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{4{x^2} + 16x + 7}}{{{{\left( {x + 2} \right)}^2}}}.\end{array}\)

Cho \(y' = 0 \Leftrightarrow \dfrac{{4{x^2} + 16x + 7}}{{{{\left( {x + 2} \right)}^2}}} = 0 \Leftrightarrow 4{x^2} + 16x + 7 = 0 \Leftrightarrow \left[ \begin{array}{l}x =  - \dfrac{1}{2}\,\,\,\,\left( {L{\rm{oai}}} \right)\\x =  - \dfrac{7}{2}\,\,\,\,\left( {Loai} \right)\end{array} \right.\)

Thay \(x = 0\)vào \(f\left( x \right)\) ta có \(f\left( 0 \right) = \dfrac{7}{2}.\)

Thay \(x = 2\)vào \(f\left( x \right)\) ta có \(f\left( 2 \right) = \dfrac{{37}}{4}.\)

\( \Rightarrow \,\max f\left( x \right) = \dfrac{{37}}{4};\,\,\min f\left( x \right) = \dfrac{7}{2}.\)