Câu hỏi
Tìm GTLN , GTNN của các hàm số sau: \(y = f(x) = x + \dfrac{9}{x}\) trên \(\left[ {2;4} \right]\)
- A \( \max f\left( x \right) =- 6;\,\,min\,f\left( x \right) = -\dfrac{{13}}{2}\).
- B \( \max f\left( x \right) = 6;\,\,min\,f\left( x \right) = -\dfrac{{13}}{2}\).
- C \( \max f\left( x \right) = \dfrac{{13}}{2};\,\,min\,f\left( x \right) =- 6\).
- D \( \max f\left( x \right) = \dfrac{{13}}{2};\,\,min\,f\left( x \right) = 6\).
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: \(f\left( 3 \right) = 6.\)
\( \Rightarrow \max f\left( x \right) = \dfrac{{13}}{2};\,\,min\,f\left( x \right) = 6\).
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