题目

y=x* e^(-x) 在x大于等于0小于等于4时的最小值

提问时间：2021-02-06

答案

y = xe^(-x),0≤x≤4
dy/dx = e^(-x) + x * (-1)e^(-x)
= e^(-x) - xe^(-x)
= (1 - x)e^(-x)
d²y/dx² = (-1)e^(-x) + (1 - x) * (-1)e^(-x)
= -e^(-x) * (1 + 1 - x)
= (x - 2)e^(-x)
dy/dx = 0 ==> 1 - x = 0 ==> x = 1
d²y/dx²|(x = 1) = -1/e < 0,取得极小值
f(1) = 1 * e^(-1) = 1/e
在端点,f(0) = 0,f(4) = 4 * e^(-4) = 4/e⁴
所以,ymin = 0