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Câu hỏi: Tìm giá trị nhỏ nhất của hàm số $y=x+\dfrac{9}{x}$ trên đoạn $\left[ 2;4 \right]$
A. $\underset{\left[ 2;4 \right]}{\mathop{\min }} y=-6$
B. $\underset{\left[ 2;4 \right]}{\mathop{\min }} y=6$
C. $\underset{\left[ 2;4 \right]}{\mathop{\min }} y=\dfrac{25}{4}$
D. $\underset{\left[ 2;4 \right]}{\mathop{\min }} y=\dfrac{13}{2}$
A. $\underset{\left[ 2;4 \right]}{\mathop{\min }} y=-6$
B. $\underset{\left[ 2;4 \right]}{\mathop{\min }} y=6$
C. $\underset{\left[ 2;4 \right]}{\mathop{\min }} y=\dfrac{25}{4}$
D. $\underset{\left[ 2;4 \right]}{\mathop{\min }} y=\dfrac{13}{2}$
Ta có $y'=1-\dfrac{9}{{{x}^{2}}}=0\Leftrightarrow \left[ \begin{aligned}
& x=3\text{ }\left( tmdk \right) \\
& x=-3 \\
\end{aligned} \right.$
$\begin{aligned}
& y\left( 2 \right)=\dfrac{13}{2} \\
& y\left( 3 \right)=6 \\
& y\left( 4 \right)=\dfrac{25}{4} \\
\end{aligned} $ $ \Rightarrow \underset{\left[ 2;4 \right]}{\mathop{\min }} y=6$
& x=3\text{ }\left( tmdk \right) \\
& x=-3 \\
\end{aligned} \right.$
$\begin{aligned}
& y\left( 2 \right)=\dfrac{13}{2} \\
& y\left( 3 \right)=6 \\
& y\left( 4 \right)=\dfrac{25}{4} \\
\end{aligned} $ $ \Rightarrow \underset{\left[ 2;4 \right]}{\mathop{\min }} y=6$
Đáp án B.