Đáp án:
\(\eqalign{
& a)\,\,\mathop {\min }\limits_{\left[ {0;1} \right]} y = - 8;\,\,\mathop {\max }\limits_{\left[ {0;1} \right]} y = - {{20} \over 3} \cr
& b)\,\,\mathop {\min }\limits_{\left[ {0;1} \right]} y = - 59;\,\,\mathop {\max }\limits_{\left[ {0;1} \right]} y = - 7 \cr} \)
Giải thích các bước giải:
\(\eqalign{
& a)\,\,y = - 3{x^2} + 4x - 8 \cr
& y' = - 6x + 4 = 0 \Leftrightarrow x = {2 \over 3} \in \left[ {0;1} \right] \cr
& y\left( 0 \right) = - 8;\,\,y\left( 1 \right) = - 7;\,\,y\left( {{2 \over 3}} \right) = - {{20} \over 3} \cr
& \Rightarrow \mathop {\min }\limits_{\left[ {0;1} \right]} y = - 8;\,\,\mathop {\max }\limits_{\left[ {0;1} \right]} y = - {{20} \over 3} \cr
& b)\,\,y = {x^3} + 3{x^2} + 9x - 7 \cr
& y' = 3{x^2} + 6x + 9 > 0\,\,\forall x \cr
& \Rightarrow \mathop {\min }\limits_{\left[ {0;1} \right]} y = y\left( { - 4} \right) = - 59;\,\,\mathop {\max }\limits_{\left[ {0;1} \right]} y = y\left( 0 \right) = - 7 \cr} \)
