jijycao421
幼苗
共回答了22个问题采纳率:90.9% 举报
f(x)=x^2.e^x
f'(x) = (2x+x^2)e^x
f'(1) = 3e
f(1) = e
图象在点(1,f(1))处的切线方程
y-f(1) = f'(1) ( x-1)
y-e = 3e(x-1)
f'(x) = (2x+x^2)e^x =0
x^2+2x=0
x=0 or -2
f''(x) =(2+ 4x+x^2)e^x
f''(0)= 2>0 ( min )
f''(-2) = -2e^(-2)
1年前
7