58.已知函数f(x)=a(x−1)ex−x2+1f(x)=a(x-1)e^{x}-x^{2}+1f(x)=a(x−1)ex−x2+1.(1)讨论f(x)f(x)f(x)的单调性;(2)若a=e−4a=e^{-4}a=e−4,证明:当x>0x>0x>0时,f(x+1)⩾lnx−x2−x−2f(x+1)\geqslant lnx-x^{2}-x-2f(x+1)⩾lnx−x2−x−2.