f(x)=ln(ax+1)+2/（x+1）
定义域要求: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时,2/a-10 ==> x∈ (√(2/a-1) ,+∞ )
f(x)增区间(√(2/a-1) ,+∞ ),减区间(0,√(2/a-1) )
②
若函数f(x)在[0,+∞）上最小值为2,求正数a的取值范围
a≥2时,2/a-1