【高中数学=导数】已知函数 .（1）试讨论f(x)的单调性；

g(x)=f(x)+(a-4)lnx+3ax-(3a+1)/x
=(2-a)lnx-2ax-1/x+(a-4)lnx+3ax-(3a+1)/x
=-2lnx+ax-(3a+2)/x
g'(x)=-2/x+a+(3a+2)/x^2=(ax^2-2x+3a+2)/x^2
g(x)在[1,4]上不单调,则说明g'(x)=0在区间上有零点.
即有ax^2-2x+3a+2=0在区间上有零点.
即有a(x^2+3)=2x-2
a=(2x-2)/(x^2+3)
设h(x)=(2x-2)/(x^2+3).则有h'(x)=[2(x^2+3)-(2x-2)*(2x)]/(x^2+3)^2=(-2x^2+4x+6)/(x^2+3)^2=0
x^2-2x-3=0
(x-3)(x+1)=0
x1=3,x2=-1
故在[1,3]上有h'(x)>0,h(x)单调增,在[3,4]上有h'(x)