f(x)=(1+x)^2+ln(1+x) (1)求f(x)单调区间 (2)若x ∈[1/e-1,e-1]时不等式f(x)

题目

f(x)=(1+x)^2+ln(1+x) (1)求f(x)单调区间 (2)若x ∈[1/e-1,e-1]时不等式f(x)

答案

f'(x)=2(x+1)(x+1)' + 1/(1+x)(x+1)'=2(x+1)+1/(x+1) (定义域：x≠-1)
（1）A：若f'(x)≥0,即2(x+1)+1/(x+1)≥0,解得x>-1
B：若f'(x)≤0,即2(x+1)+1/(x+1)≤0,解得x-1,所以x ∈[1/e-1,e-1],在f(x)的单调递增区间上,
所以,f(x)≤f(e-1),故不等式f(x)