Question

！已知函数f（x）＝a ln x-ax-3(a∈r 且a≠0).求函数f（x）的单调区间

Answer 1

f(x)=ln(2-x)
ax,
定义域:2-x>0,
x<2
∴f'(x)=-1/(2-x)
a=a-1/(2-x)
令f'(x)>=0,
则a>=1/(2-x),
2-x>=1/a,
∴x<=2-1/a
令f'(x)<=0,
则a<=1/(2-x),
2-x<=1/a,
∴x>=2-1/a
综上,单调增区间为(-∞,2-1/a],单调减区间为[2-1/a,2)