f(x)=ln(ax+1)+2/（x+1）
定義域要求:x>0,且x>-1/a
f'(x)=a/(ax+1)-2/(x+1)^2
=[a(x+1)^2-2(ax+1)]/[(ax+1)(x+1)^2]
=(ax^2+a-2)/[(ax+1)(x+1)^2]
=a[x^2-(2/a-1)]/[(ax+1)(x+1)^2]
a≥2時(shí),2/a-10 ==> x∈ (√(2/a-1) ,+∞ )
f(x)增區(qū)間(√(2/a-1) ,+∞ ),減區(qū)間(0,√(2/a-1) )
②
若函數(shù)f(x)在[0,+∞）上最小值為2,求正數(shù)a的取值范圍
a≥2時(shí),2/a-1