答案:-5/12

6.(2018·长春模拟)已知函数f(x)={■(x+2/e "," x<0"," @x/e^x "," x≥0"," )┤若f(x1)=f(x2)= f(x3)(x1

【解题指南】利用导数法,分析函数的单调性及极值,可得f(x1)=f(x2)=f(x3)∈(0"," 1/e),即有-2/e

【解析】函数f(x)={■(x+2/e "," x<0"," @x/e^x "," x≥0"," )┤

所以函数f'(x)={■(1"," x<0"," @(1"-" x)/e^x "," x≥0"," )┤

故当x<0时,函数是增加的,且f(x)<2/e,

当0≤x<1时,函数是增加的,且0≤f(x)<1/e,

当x≥1时,函数是减少的,且0

若f(x1)=f(x2)=f(x3)(x1

则f(x1)=f(x2)=f(x3)∈(0"," 1/e),

即-2/e

故(f"(" x_2 ")" )/x_1 =(f"(" x_1 ")" )/x_1 =1+2/(ex_1 )∈(-1,0).

答案:(-1,0)

三、解答题(每小题10分,共20分)

7.已知函数f(x)=(e^x+2)/x. 导学号