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),
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