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Câu hỏi: Tìm giá trị lớn nhất của hàm số $f\left( x \right)=2{{x}^{3}}+3{{x}^{2}}-12x+2$ trên đoạn $\left[ -1;2 \right].$
A. $\underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=6.$
B. $\underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=10.$
C. $\underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=15.$
D. $\underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=-5$
A. $\underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=6.$
B. $\underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=10.$
C. $\underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=15.$
D. $\underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=-5$
Đạo hàm $f'\left( x \right)=6{{x}^{2}}+6x-12\xrightarrow{{}}f'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=1\in \left[ -1;2 \right] \\
& x=-2\notin \left[ -1;2 \right] \\
\end{aligned} \right..$
Ta có $\left\{ \begin{aligned}
& f\left( -1 \right)=15 \\
& f\left( 1 \right)=-5 \\
& f\left( 2 \right)=6 \\
\end{aligned} \right.\xrightarrow{{}}\underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=15.$
& x=1\in \left[ -1;2 \right] \\
& x=-2\notin \left[ -1;2 \right] \\
\end{aligned} \right..$
Ta có $\left\{ \begin{aligned}
& f\left( -1 \right)=15 \\
& f\left( 1 \right)=-5 \\
& f\left( 2 \right)=6 \\
\end{aligned} \right.\xrightarrow{{}}\underset{\left[ -1;2 \right]}{\mathop{\max }} f\left( x \right)=15.$
Đáp án C.