解: (Ⅰ) 由 x=0是极值点,故,得 故 m=1. 故 当 -1<x<0时,函数在(-1,0)内是减函数; 当 x>0时,函数f(x)在(0,+∞)内是增函数。 所以x=0时,f(0)=0,则函数f(x)取得最小值为0.····························6分 (Ⅱ)由(Ⅰ)知:f(x)≥0,故ex-1≥ln(x+1)。 ∵①··············8分 又 = 故 ················································10分 故 ② 由①②得 ················12分 |