已知f0(x)=xe^x,定义fn(x)=f'(n-1)(x) x属于N,试归纳出fn（x）的表达式
问题描述：
已知f0(x)=xe^x,定义fn(x)=f'(n-1)(x) x属于N,试归纳出fn（x）的表达式
求fn(x)的极小值，点Pn(Xn,yn)
答
f0(x) = xe^x
f1(x) = f'(0)(x)
=(x+1)e^x
f2(x) = f'(1)x
= (x+2)e^x
...
...
fn(x) = f'(n-1)x
= (x+n)e^x求fn(x)的极小值，点Pn(Xn,yn)求下这个吧...说实话，我是没看懂- -fn(x) = (x+n)e^xf'n(x) = (x+n+1)e^x =0x= -(n+1)f''n(x) = (x+n+2)e^xf''n(-(n+1)) = e^(-(n+1)) >0 ( min )min fn(x) = fn(-n+1)) = -e^(-(n+1))Pn(xn,yn) = (-(n+1), -e^(-(n+1)))3Q