Đáp án:
\(\left[ \matrix{
m = 2 \hfill \cr
m = - {7 \over 6} \hfill \cr} \right.\)
Giải thích các bước giải:
\(\eqalign{
& y = {x^3} - \left( {{m^2} + 2} \right){x^2} + \left( {5m - 1} \right)x - 1 \cr
& y' = 3{x^2} - 2\left( {{m^2} + 2} \right)x + 5m - 1 \cr
& y'' = 6x - 2\left( {{m^2} + 2} \right) \cr
& Ham\,\,so\,\,dat\,\,cuc\,\,tieu\,\,tai\,\,x = 3 \cr
& \Rightarrow \left\{ \matrix{
y'\left( 3 \right) = 0 \hfill \cr
y''\left( 3 \right) > 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
27 - 6\left( {{m^2} + 2} \right) + 5m - 1 = 0 \hfill \cr
18 - 2\left( {{m^2} + 2} \right) > 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
- 6{m^2} + 5m + 14 = 0 \hfill \cr
{m^2} + 2 < 9 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
\left[ \matrix{
m = 2 \hfill \cr
m = - {7 \over 6} \hfill \cr} \right. \hfill \cr
- \sqrt 7 < m < \sqrt 7 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
m = 2 \hfill \cr
m = - {7 \over 6} \hfill \cr} \right. \cr} \)
