Question

设函数f（x）=（x²+ax）e^kx（a，k∈R），已知f（x）在区间（-∞，-］?和［，+∞）上单调递增，在区间（-，）上单调递减.

（1）求a和k的值；

（2）求函数f（x）在区间［0，m］（m＞0）上的最大值和最小值.

Answer 1

解:(1)∵f′(x)=［kx²+(ak+2)x+a］e^kx,                                                           ?

而e^kx＞0,∴由题意知方程kx²+(ak+2)x+a=0的两根为-,.?

∴                                                                                      ?

解得或(舍).                                                                                 ?

(2)∵f(x)在(-∞,-］和［,+∞)上递增,在［-,］上递减,

且f(0)=0,f(2)=0,f(m)=(m²-2m)e^m.?

(ⅰ)当0＜m≤2时,［f(x)］_Min=f(m)=(m²-2m)e^m,［f(x)］_Max=f(0)=0.                         ?

(ⅱ)当＜m≤2时,［f(x)］_Min=f()=(2-2),［f(x)］_Max=f(0)=0.        ?

(ⅲ)当m＞2时,［f(x)］_Min=f()=(2-2),［f(x)］_Max=f(m)=(m²-2m)e^m.