答案:解:先求驻点,由[tex=12.571x3.5]7EJHVCtO2IWq3KpdB+jQsj7pfeJjzwos6SIkSNru0jeb5fSCm94gzwUV3pr2dvRTX5PqzvKnA8OPIVE9I9uxtM2LSeHYvHT/ZkoHjBIT3u5JxNAKx434fRAIChs/5JnR/ykfzFPLdKCyRNlkUkG6BdaKV78QQbstuHbTZdUg5HI=[/tex],解得[tex=16.286x2.643]9WsicLRp/TPdMYxedjgTNiuyOHJBD9cNd0aECY0AgTWcVeIltpR7y7+36zo+dqXXbm34eC0YNS+xcWOzfyorFpPr7qiyzRmn8mQVKH4on6Q=[/tex],即驻点为[tex=13.571x3.357]uxka8Mx4Js74hQb06lbah75Sl/fiZfCn0aF8eCM3OrIEOOMCvLffIfKUicfp2oxW+sX/pCo21ROmu/aSm1KXYrTHATQFMNq8yyGCczZ+t3A=[/tex]和[tex=5.643x3.357]JH6dQwirShE8V9E13B0yMP+rQRtgaAkxAPAL3R7Gqa18Yz21xhIehrAvtoiG3r+q[/tex]五点,再由[tex=18.5x1.5]SwipKkCVeKtuPilERL6UvKzG8xioVLcuFjy7PE6Gcg3xOC6PKpeJbSnneVe00c350/hOLQv9MGgF6wfvlQu8COqotQFSLYo9Atdb7pBqJyc=[/tex]可知[tex=15.929x1.714]3DddgVP1x+/PDfBvnEh3mcOpcua5Zt2UczSe8lRV7/CD9a/1u4tM1GDXVz7UI8On8UJTwdAp5py6PmwhuUoBFbASsmG4xLS/6gzX8m2mYz8=[/tex]。应用定理12.6.2驻点[tex=11.071x3.357]lyDK4d4EHIxGE0JtmUUs1rgS2lXJL/X6XtbzD6yyTkl5FBpidr85eX+G4tXobwZbLALxBVYRXJ+tokb0XOB/LoA/BnmVhKDDhjwx4UhIjbXJdttEVgZiQUWtZ1BZozXA[/tex]满足H>0,所以是极值点,再根据[tex=0.929x1.214]rptEmx/7E1A4KUr49a04Qw==[/tex]的符号,可知函数在[tex=11.071x3.357]lyDK4d4EHIxGE0JtmUUs1rgS2lXJL/X6XtbzD6yyTknm5DO0lHHCkF9oDBjBbRprCREGayyDdRWknEiVx6xe0bIDzHTSRM8CTrghwf45T9AxcZTKON9lf8FByR0xUlgo[/tex]取极小值[tex=2.214x2.357]NbIZsEQ2IM+2V5opmeY28Q==[/tex]。在 (1,1),(-1,1) 点 H <0, 所以 (1,1),(-1,1) 不是极值点。在 (0,0) 点 H =0, 且 f (0,0)=0 .由于[tex=9.286x1.571]V0QxJSKklaH1qoyUG/7CgDQOjZeVUXWheWctw/4egECrwDG7Pn7KHl+rKOA9t5OL[/tex], 易知函数在 (0,0) 点附近变化,所以 (0,0)不是极值点.