Ta có: \(f'(x) = 1 - \dfrac{9}{{{x^2}}} = \dfrac{{{x^2} - 9}}{{{x^2}}}\)
\(f'(x) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 3 \in \left[ {2;4} \right]\\x = - 3 \notin \left[ {2;4} \right]\end{array} \right.\)
Mà \(f\left( 2 \right) = \dfrac{{13}}{2},f\left( 3 \right) = 6,f\left( 4 \right) = \dfrac{{25}}{4}\)
Suy ra : \(\mathop {\min }\limits_{{\rm{[}}2;4]} f(x) = 6;\mathop {\max }\limits_{{\rm{[}}2;4]} f(x) = \dfrac{{13}}{2}\).
