Question

设函数f(x)=ax-(a+1)ln(x+1),(a>0)

(1)求fx的单调区间（2）求fx在区间【1，2】上的最小值

Answer 1

f(x)=ax-(a+1)ln(x+1)f'(x)=a-(a+1)/(x+1)=[a(x+1)-(a+1)]/(x+1)=(ax-1)/(x+1)1)定义域:x+1>0, ∴x>-1令f'(x)>=0, 则x>=1/a令f'(x)<=0, 则-1<x<=1/a∴f(x)单调增区间为[1/a,+∞),单调减区间为(-1,1/a]2)若1/a>=2,即0<a<=1/2,此时f(x)在[1,2]上单调减∴最小值为 f(2)=2a-(a+1)ln3若1/a<=1,即a>=1,此时f(x)在[1,2]上单调增∴最小值为 f(1)=a-(a+1)ln2若1<1/a<2,即1/2<a<1,此时f(x)在[1,1/a]单调减,在[1/a,2]上单调增∴最小值为 f(1/a)=1-(a+1)ln(1/a+1)综上,0<a<=1/2时为2a-(a+1)ln3; 1/2<a<1时为1-(a+1)ln(1/a+1); a>=1时为a-(a+1)ln2