tìm giá trị lớn nhất, giá trị nhỏ nhất của hàm số: a) y= x^3-3x^2-9x+35 trên các đoạn [-4;4] và [0;5] b)y=căn 5-4x trên đoạn [-1;1]

$$\eqalign{ & a)\,\,y = {x^3} - 3{x^2} - 9x + 35\,\,\,\,\,tren\,\,\left[ { - 4;4} \right] \cr & y' = 3{x^2} - 6x - 9 = 0 \Leftrightarrow \left[ \matrix{ x = 3 \in \left[ { - 4;4} \right] \hfill \cr x = - 1 \in \left[ { - 4;4} \right] \hfill \cr} \right. \cr & y\left( { - 4} \right) = - 41;\,\,y\left( 4 \right) = 14;\,\,y\left( 3 \right) = 8;\,\,y\left( { - 1} \right) = 40 \cr & \Rightarrow \mathop {\min }\limits_{\left[ { - 4;4} \right]} y = - 41;\,\,\mathop {\max }\limits_{\left[ { - 4;4} \right]} y = 40 \cr & b)\,\,y = \sqrt {5 - 4x} \,\,xac\,\,dinh\,\,tren\left[ { - 1;1} \right] \cr & y' = {{ - 4} \over {2\sqrt {5 - 4x} }} < 0\,\,\forall x \in \left[ { - 1;1} \right] \cr & y\left( { - 1} \right) = 3;\,\,y\left( 1 \right) = 1 \cr & \Rightarrow \mathop {\min }\limits_{\left[ { - 1;1} \right]} y = 1;\,\,\mathop {\max }\limits_{\left[ { - 1;1} \right]} y = 3 \cr} $$

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