f(x)=x(12-2x)^2 trên [0,6] tìm min max

Đáp án:

$\begin{array}{l}
f\left( x \right) = x{\left( {12 - 2x} \right)^2} = 4.x.{\left( {6 - x} \right)^2}\\
 = 4.x.\left( {{x^2} - 12x + 36} \right)\\
 = 4{x^3} - 48{x^2} + 144x\\
 \Rightarrow f'\left( x \right) = 12{x^2} - 96x + 144 = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 6
\end{array} \right.\\
Do:\left\{ \begin{array}{l}
f\left( 0 \right) = 0\\
f\left( 2 \right) = 128\\
f\left( 6 \right) = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
\min \,f\left( x \right) = 0\\
max\,f\left( x \right) = 128
\end{array} \right.
\end{array}$