设f1(x)=x^2+mx+m,f2(x)=e^-x则f(x)=f1(x)*f2(x)m≤2,f1(x)的对称轴为x=-m/2≤-1所以f1(x)在[-1,+∞）上为增函数e=2.71828＞1,x≥0所以f2(x)在R上为减函数所以f(x)在[0,+∞）上为减函数因为f(0)=m≤2所以当x>=0时,f(...
