Question

设x＝1和x＝2是函数f(x)＝aln＋bx ＋x,的两个极值点,求a.b的值,求f(x)的单调区间

Answer 1

f(x)=alnx+bx^2+x是这样的吧.

f'(x)=a/x+2bx+1

x=1,x=2是极值点,则有f'(1)=a+2b+1=0

f'(2)=a/2+4b+1=0

解得:a=-2/3,b=-1/6

f(x)=-2/3lnx-1/6x^2+x

f'(x)=-(2/3)/x-1/3x+1=-(x^2-3x+2)/(3x)=-(x-1)(x-2)/(3x)>0

得到1