Đáp án:
\(\begin{array}{l}
\mathop {\min y}\limits_{{\rm{[}} - 1,1]} = y(\frac{{ - 1 + \sqrt 3 }}{2}) = \frac{{ - 1}}{4}\\
\mathop {\max y}\limits_{{\rm{[}} - 1,1]} = y(1) = 2
\end{array}\)
Giải thích các bước giải:
y'=4x³+6x²-1=0 <-> \(\left[ \begin{array}{l}
x = \frac{{ - 1 - \sqrt 3 }}{2} \notin {\rm{[}} - 1,1]\\
x = \frac{{ - 1 + \sqrt 3 }}{2} \in {\rm{[}} - 1,1]\\
x = \frac{{ - 1}}{2} \in {\rm{[}} - 1,1]
\end{array} \right.\)
-> \(\begin{array}{l}
y( - 1) = 0\\
y(1) = 2\\
y(\frac{{ - 1 + \sqrt 3 }}{2}) = \frac{{ - 1}}{4}\\
y(\frac{{ - 1}}{2}) = \frac{5}{{16}}
\end{array}\)
-> \(\begin{array}{l}
\mathop {\min y}\limits_{{\rm{[}} - 1,1]} = y(\frac{{ - 1 + \sqrt 3 }}{2}) = \frac{{ - 1}}{4}\\
\mathop {\max y}\limits_{{\rm{[}} - 1,1]} = y(1) = 2
\end{array}\)
