引例已知函数f(x)=xe~x+c有两个零点,则实数c的取值范围是.错解令f(x)=xe~x+c=0,得xe~x=-c.记g(x)=xe~x,则g′(x)=e~x(x+1),令g′(x)=0,得x=-1.当x<-1时,g′(x)<0,g(x)单调递减;当x>-1时,g′(x)>0,g(x)单调递增.于是,函数g(x)的图象大致如图1所示.要使f(x)有两个零点,即使直线y=-c与g(x)的图象有两个不同的交点,结合图1可知:有 For instance, if the function f (x) = xe ~ x + c has two zeroes, then the range of the real number c is 0. The solution f (x) = xe ~ x + c = Let g ’(x) = e ~ x (x + 1), let g’ (x) = 0, so that x = -1. When x <-1, g ’(x) <0, g (x) decreases monotonically; when x> -1, g’ (x)> 0, g 1. To make f (x) have two zeroes, even though there are two different intersections between the images of the line y = -c and g (x), we can see from Figure 1 that there is