Question

设x=1与x=2是函数f(x)=alnx+bx²+x的两个极值点（1）试确定常数a和b的值

Answer 1

根据题意有f'(x)=0的解为x=1或x=2

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

-1/2b=1+2=3=>b=-1/6

a/2b=1*2=2=>a=-2/3

所以f'(x)=[(-1/3)x^2+x-2/3]/x=-(x-1)(x-2)/3x

(-∞,0)f'(x)>0,单增

(0,1]f'(x)