Question

已知函数f(x)=(k-x)ex,求f(x)在区间[0,2]上的最大值

Answer 1

f(x)=(k-x)e^x

f'(x)=-e^x+(k-x)e^x=(k-1-x)e^x=0

得

x=k-1

1.k-1属于[0,2]

则

由f''(x)=(k-2-x)e^x

得

f''(k-1)=-e^(k-1)2,k>3

f'(x)=(k-1-x)e^x>0

f(x)递增,最大值=f(2)=(k-2)e^2

3.k-1