(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