Question

已知函数f(x)=x^2+lnx-ax(a>0),求f(x)的单调区间

Answer 1

首先,保证真数大于0,有x>0

对f(x)求导得

f’(x)=2x+1/x-a=(2x²-ax+1)/x

令f’(x)=0得(2x²-ax+1)/x=0,即2x²-ax+1=0,再令其判别式△=a²-4=0得a=2

1、当02时,△>0,令f’(x)≥0以求原函数的增区间,得(2x²-ax+1)/x≥0,因为x>0,不等式两边同乘以x,不等式不变号得2x²-ax+1≥0,解之得

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