Đáp án:
$\begin{array}{l}
1)y = {x^3} - 3{x^2} - 9x + 35\\
\Rightarrow y' = 3{x^2} - 6x - 9 = 0\\
\Rightarrow 3\left( {{x^2} - 2x - 3} \right) = 0\\
\Rightarrow 3\left( {x - 3} \right)\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{y_{ - 4}} = - 41\\
{y_{ - 1}} = 40\\
{y_3} = 8\\
{y_4} = 15
\end{array} \right.\\
\Rightarrow GTLN:y = 40\,khi:x = - 1\\
C2)\\
y = {x^3} - 7{x^2} + 11x - 2\\
\Rightarrow y' = 3{x^2} - 14x + 11 = 0\\
\Rightarrow \left( {3x - 11} \right)\left( {x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{{11}}{3}\left( {ktm:x \in \left[ {0;2} \right]} \right)\\
x = 1\left( {tm} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{y_0} = - 2\\
{y_1} = 3\\
{y_2} = 0
\end{array} \right.\\
\Rightarrow GTNN:y = - 2\,khi:x = 0
\end{array}$
