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Câu hỏi: Giá trị lớn nhất của hàm số $f\left( x \right)={{x}^{3}}-8{{x}^{2}}+16x-9$ trên đoạn $\left[ 1;3 \right]$ là
A. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=5.$
B. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=-6.$
C. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=\dfrac{13}{27}.$
D. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=0.$
A. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=5.$
B. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=-6.$
C. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=\dfrac{13}{27}.$
D. $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=0.$
Hàm số liên tục trên đoạn [1; 3].
+ Ta có: $f'\left( x \right)=3{{x}^{2}}-16x+16;f'\left( x \right)=0\Leftrightarrow 3{{x}^{2}}-16x+16=0\Leftrightarrow \left[ \begin{aligned}
& x=4\notin \left[ 1;3 \right] \\
& x=\dfrac{4}{3}\in \left[ 1;3 \right] \\
\end{aligned} \right.$
+ $f\left( 1 \right)=0;f\left( 3 \right)=-6;f\left( \dfrac{4}{3} \right)=\dfrac{13}{27}.$ Vậy $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=\dfrac{13}{27}.$
+ Ta có: $f'\left( x \right)=3{{x}^{2}}-16x+16;f'\left( x \right)=0\Leftrightarrow 3{{x}^{2}}-16x+16=0\Leftrightarrow \left[ \begin{aligned}
& x=4\notin \left[ 1;3 \right] \\
& x=\dfrac{4}{3}\in \left[ 1;3 \right] \\
\end{aligned} \right.$
+ $f\left( 1 \right)=0;f\left( 3 \right)=-6;f\left( \dfrac{4}{3} \right)=\dfrac{13}{27}.$ Vậy $\underset{\left[ 1;3 \right]}{\mathop{\max }} f\left( x \right)=\dfrac{13}{27}.$
Đáp án C.