∵g(x)与f(x)关于x=1对称∴g(1+x)=f(1-x)=(1-x)e^[-(1-x)]=(1-x)e^(x-1)令t=1+xx=t-1g(t)=[1-(t-1)]e^[(t-1)-1]=(2-t)e^(t-2)把t换成xg(x)=(2-x)e^(x-2)
g(x)=f(2-x)=代入就可以
