(1)f'(x)=(x^2+2x)·[e^(x-1)-1]
导数大于0时
解得x>1,x∈（-2,0）
导数小于0时
解xx因为对于函数f(x)= e^(x-1)-x,f'(x)=e^(x-1)-1, 当x>1时,f'(x)>0,所以当x>1时f(x)= e^(x-1)-x为增函数,f(1)=0 所以 当x>1时f(x)>0e^(x-1)>x
2 设n=k时, e^(x-1)>x ^k/k!
因为x^k/k!-x^（k+1）/（k+1）!=x^k/k!-x^k/k![x/(k+1)]=x^k/k![1-x/(k+1)]=x^k/k![(k+1-x)/k+1)]
因为x可看作常数,当k足够大时必有k>x所以(k+1-x)/(k+1)>0,所以x^k/k!-x^（k+1）/（k+1）!>0,所以x^k/k!>x^（k+1）/（k+1）!所以 e^(x-1)>x^（k+1）/（k+1）!
所以当n=k+1时原式成立,所以 e^(x-1)>x ^n/n!n>=1,n是整数 x>1