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Câu hỏi: Tìm giá trị lớn nhất của hàm số $f\left( x \right)={{x}^{3}}-8{{x}^{2}}+16x-9$ trên đoạn $\left[ 1; 3 \right]$
A. $\underset{x\in \left[ 1; 3 \right]}{\mathop{\max }} f\left( x \right)=-6$.
B. $\underset{x\in \left[ 1; 3 \right]}{\mathop{\max }} f\left( x \right)=\dfrac{13}{27}$.
C. $\underset{x\in \left[ 1; 3 \right]}{\mathop{\max }} f\left( x \right)=5$.
D. $\underset{x\in \left[ 1; 3 \right]}{\mathop{\max }} f\left( x \right)=0$.
A. $\underset{x\in \left[ 1; 3 \right]}{\mathop{\max }} f\left( x \right)=-6$.
B. $\underset{x\in \left[ 1; 3 \right]}{\mathop{\max }} f\left( x \right)=\dfrac{13}{27}$.
C. $\underset{x\in \left[ 1; 3 \right]}{\mathop{\max }} f\left( x \right)=5$.
D. $\underset{x\in \left[ 1; 3 \right]}{\mathop{\max }} f\left( x \right)=0$.
Ta có: ${f}'\left( x \right)=3{{x}^{2}}-16x+16\Rightarrow {f}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=\dfrac{4}{3}\in \left( 1;3 \right) \\
& x=4\notin \left( 1;3 \right) \\
\end{aligned} \right.$
Và: $f\left( 1 \right)=0,f\left( \dfrac{4}{3} \right)=-\dfrac{115}{27},f\left( 3 \right)=-6\Rightarrow \underset{x\in \left[ 1; 3 \right]}{\mathop{\max }} f\left( x \right)=0$
& x=\dfrac{4}{3}\in \left( 1;3 \right) \\
& x=4\notin \left( 1;3 \right) \\
\end{aligned} \right.$
Và: $f\left( 1 \right)=0,f\left( \dfrac{4}{3} \right)=-\dfrac{115}{27},f\left( 3 \right)=-6\Rightarrow \underset{x\in \left[ 1; 3 \right]}{\mathop{\max }} f\left( x \right)=0$
Đáp án D.