设u=ax+2a+1，当-1≤x≤1时，u的值有正有负

当x=-1,1时
u=-a+2a+1,即a+1
u=a+2a+1,即3a+1
因为u有正有负，
所以有两种情况
1.a+1＞0，3a+1<0 得-1<a<-1/3
2a+1<0，3a+1＞0 得-1>a>-1/3,错误，舍去
即-1<a<-1/3

-1< a < -1/3

x = (u-2a - 1)/ a( a!= 0)
-1 <= (u-2a - 1)/ a <= 1
如果 a > 0,
-a <= (u-2a - 1) <= a
a + 1 <= u <= 3a + 1;
a+1 < 0 and 3a + 1 > 0 and a > 0
so no answer.

then a < 0,
-a >= (u - 2a - 1 ) >= a
a + 1 >= u >= 3a + 1
a + 1 > 0 and 3a + 1 < 0 and a < 0;
so -1 < a < -1/3

u'=a 有题意得 u为单调函数

u（-1）=a+1

u（1）=3a+1

-1<a<-1/3