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Câu hỏi: Tìm giá trị lớn nhất của hàm số $f\left( x \right)={{x}^{3}}-2{{x}^{2}}-4x+1$ trên đoạn $\left[ 1;3 \right].$
A. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=\dfrac{67}{27}.$
B. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=-2.$
C. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=-7.$
D. $x\left( \text{cm} \right)$
A. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=\dfrac{67}{27}.$
B. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=-2.$
C. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=-7.$
D. $x\left( \text{cm} \right)$
Đạo hàm $f'\left( x \right)=3{{x}^{2}}-4x-4\xrightarrow{{}}f'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=2\in \left[ 1;3 \right] \\
& x=-\dfrac{2}{3}\notin \left[ 1;3 \right] \\
\end{aligned} \right..$
Ta có $\left\{ \begin{aligned}
& f\left( 1 \right)=-4 \\
& f\left( 2 \right)=-7 \\
& f\left( 3 \right)=-2 \\
\end{aligned} \right.\xrightarrow{{}}\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=-2.$
& x=2\in \left[ 1;3 \right] \\
& x=-\dfrac{2}{3}\notin \left[ 1;3 \right] \\
\end{aligned} \right..$
Ta có $\left\{ \begin{aligned}
& f\left( 1 \right)=-4 \\
& f\left( 2 \right)=-7 \\
& f\left( 3 \right)=-2 \\
\end{aligned} \right.\xrightarrow{{}}\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=-2.$
Đáp án B.