3.(2018·全国卷Ⅲ)已知函数f(x)=(ax^2+x"-" 1)/e^x .

(1)求曲线y=f(x)在点(0",-" 1)处的切线方程.

(2)证明:当a≥1时,f(x)+e≥0.

【解析】(1)f(x)的定义域为R,

f'(x)=("-" ax^2+"(" 2a"-" 1")" x+2)/e^x ,

显然f(0)=-1,即点(0,-1)在曲线y=f(x)上,

所求切线斜率为k=f'(0)=2,

所以切线方程为y-(-1)=2(x-0),

即2x-y-1=0.

(2)方法一(一边为0):令g(x)=-ax2+(2a-1)x+2,

当a≥1时,方程g(x)的判别式Δ=(2a+1)2>0,

由g(x)=0得,x=-1/a,2,且-1/a<0<2,

x,f'(x),f(x)的关系如表:

x ("-∞,-" 1/a) -1/a ("-" 1/a "," 2) 2 (2,+∞) f'(x) - 0 + 0 -