Phương pháp giải:

Lời giải chi tiết:

+ TXĐ: \(D = \left[ {2;4} \right]\)

+ \(y' = f'\left( x \right) = 1 - \dfrac{9}{{{x^2}}}.\)

Cho \(y' = 0 \Leftrightarrow \,1 - \dfrac{9}{{{x^2}}} = 0 \Leftrightarrow \left[ \begin{array}{l}x = 3\,\,\,\,\,\left( {tm} \right)\\x =  - 3\,\left( {Loai} \right)\end{array} \right.\)

Thay \(x = 2\)vào \(f\left( x \right)\)ta có: \(f\left( 2 \right) = \dfrac{{13}}{2}.\)

Thay \(x = 4\)vào \(f\left( x \right)\)ta có: \(f\left( 4 \right) = \dfrac{{25}}{4}.\)

Thay \(x = 3\)vào \(f\left( x \right)\)ta có: \(f\left( 3 \right) = 6.\)

\( \Rightarrow \max f\left( x \right) = \dfrac{{13}}{2};\,\,min\,f\left( x \right) = 6\).