Câu hỏi 5

\(y={{x}^{4}}-{{x}^{2}}+13\Rightarrow y'=4{{x}^{3}}-2x\)

\(\begin{array}{l}y' = 0 \Leftrightarrow 4{x^3} - 2x = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0 \in \left[ { - 2;3} \right]\\x =  \pm \frac{1}{{\sqrt 2 }} \in \left[ { - 2;3} \right]\end{array} \right.\\y\left( { - 1} \right) = 13;y\left( 3 \right) = 85;y\left( 0 \right) = 13;y\left( { \pm \frac{1}{{\sqrt 2 }}} \right) = \frac{{51}}{4}\end{array}\)

Vậy, \(\underset{\left[ -1;3 \right]}{\mathop{Min}}\,y=y\left( \pm \frac{1}{\sqrt{2}} \right)=\frac{51}{4}\).

Chọn: A