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Câu hỏi: Tìm giá trị lớn nhất của hàm số $y=f\left( x \right)={{x}^{3}}-2{{x}^{2}}+x-2$ trên đoạn $\left[ 0;2 \right]$.
A. $\underset{\left[ 0;2 \right]}{\mathop{max}} y=-2$.
B. $\underset{\left[ 0;2 \right]}{\mathop{max}} y=-\dfrac{50}{27}$.
C. $\underset{\left[ 0;2 \right]}{\mathop{max}} y=1$.
D. $\underset{\left[ 0;2 \right]}{\mathop{max}} y=0$.
A. $\underset{\left[ 0;2 \right]}{\mathop{max}} y=-2$.
B. $\underset{\left[ 0;2 \right]}{\mathop{max}} y=-\dfrac{50}{27}$.
C. $\underset{\left[ 0;2 \right]}{\mathop{max}} y=1$.
D. $\underset{\left[ 0;2 \right]}{\mathop{max}} y=0$.
${f}'\left( x \right)=3{{x}^{2}}-4x+1$
${f}'\left( x \right)=0\Leftrightarrow 3{{x}^{2}}-4x+1=0\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x=\dfrac{1}{3} \\
\end{aligned} \right.$
$f\left( 0 \right)=-2; f\left( \dfrac{1}{3} \right)=-\dfrac{50}{27}; f\left( 1 \right)=-2; f\left( 2 \right)=0$
$\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{max}} y=f\left( 2 \right)=0$
${f}'\left( x \right)=0\Leftrightarrow 3{{x}^{2}}-4x+1=0\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x=\dfrac{1}{3} \\
\end{aligned} \right.$
$f\left( 0 \right)=-2; f\left( \dfrac{1}{3} \right)=-\dfrac{50}{27}; f\left( 1 \right)=-2; f\left( 2 \right)=0$
$\Rightarrow \underset{\left[ 0;2 \right]}{\mathop{max}} y=f\left( 2 \right)=0$
Đáp án D.