已知函数f（x）=ax+x2﹣xlna（a＞0，a≠1）．
求函数f（x）单调区间
解：函数f（x）的定义域为R，f'（x）=axlna+2x﹣lna=2x+（ax﹣1）lna．
令h（x）=f'（x）=2x+（ax﹣1）lna，h'（x）=2+axln2a，
当a＞0，a≠1时，h'（x）＞0，所以h（x）在R上是增函数，
又h（0）=f'（0）=0，所以，f'（x）＞0的解集为（0，+∞），f'（x）＜0的解集为（﹣∞，0），
This is Jim. He is thirteen. He is from the USA. He is a student. He has a good friend. His name is Jack. Jack is fifteen. He is from Canada. Jack is tall but Jim is short. They are in the same class, but they are in different grades. Jim is in Class Two. Jack is in Class Two, too. They go to school(上学) at seven in the morning and go home(回家) at six in the afternoon.
Are you good ____children?
