\[\begin{array}{l}
a)\,\,y = {x^3} - 3{x^2} - 9x + 35,\,\,\,\,\left[ { - 4;\,\,4} \right];\,\,\,\left[ {0;\,\,5} \right]\\
\Rightarrow y' = 3{x^2} - 6x - 9 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y\left( { - 4} \right) = - 41\\
y\left( { - 1} \right) = 40\\
y\left( 0 \right) = 35\\
y\left( 3 \right) = 8\\
y\left( 4 \right) = 15\\
y\left( 5 \right) = 40
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\mathop {Min}\limits_{\left[ { - 4;\,\,4} \right]} y = y\left( { - 4} \right) = - 41\\
\mathop {Max}\limits_{\left[ { - 4;\,\,4} \right]} y = y\left( 4 \right) = 15\\
\mathop {Min}\limits_{\left[ {0;\,\,5} \right]} y = y\left( 3 \right) = 8\\
\mathop {Max}\limits_{\left[ {0;\,\,5} \right]} y = y\left( 5 \right) = 40
\end{array} \right.\\
b)\,\,y = \sqrt {5 - 4x} \\
TXD:\,\,\,D = \left( { - \infty ;\,\,\frac{5}{4}} \right]\\
\Rightarrow y' = \frac{{ - 4}}{{2\sqrt {5 - 4x} }} = - \frac{2}{{\sqrt {5 - 4x} }} < 0\,\,\forall x \in D\\
\Rightarrow hs\,\,\,DB\,\,\,tren\,\,\left( { - \infty ;\,\,\frac{5}{4}} \right)\\
\Rightarrow \mathop {Min}\limits_{\left[ { - 1;\,\,1} \right]} y = y\left( 1 \right) = 1\\
\mathop {\max }\limits_{\left[ { - 1;\,\,1} \right]} y = y\left( { - 1} \right) = 3.
\end{array}\]
