Tập giá trị của hàm số \(f\...

D

- Hướng dẫn giải

\(\begin{array}{l}f'\left( x \right) = 1 - \frac{9}{{{x^2}}} = \frac{{{x^2} - 9}}{{{x^2}}}\\ \Rightarrow f'\left( x \right) = 0\\ \Leftrightarrow {x^2} - 9 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 3 \in \left[ {2;4} \right]\\x = - 3 \notin \left[ {2;4} \right]\end{array} \right..\end{array}\)

Ta có:

\(\begin{array}{l}\left\{ \begin{array}{l}f\left( 2 \right) = \frac{{13}}{2}\\f\left( 3 \right) = 6\\f\left( 4 \right) = \frac{{25}}{4}\end{array} \right.\\ \Rightarrow \mathop {\min }\limits_{\left[ {2;4} \right]} f\left( x \right) = 6;\,{\rm{ }}\mathop {\max }\limits_{\left[ {2;4} \right]} f\left( x \right) = \frac{{13}}{2}\end{array}\)

\(\begin{array}{l} \Rightarrow \left[ {a;b} \right] = \left[ {6;\frac{{13}}{2}} \right]\\ \Rightarrow P = b - a = \frac{{13}}{2} - 6 = \frac{1}{2}.\end{array}\)