case 1:
1-2x-3x^2 > 0
3x^2+2x-1<0
(3x-1)(x+1)<0
-1
case 2:
1+x 不等于0
x不等于-1
and
case 3:
-1<=2x <= 1
-1/2 <=x <=1/2
定义域
case 1 and case 2 and case 3
-1
f(x)=arcsin 2x/(1+x) + √(1-2x-3x^2)的定义域是?
f(x)=arcsin 2x/(1+x) + √(1-2x-3x^2)的定义域是?
数学人气：745 ℃时间：2020-05-21 13:42:57
优质解答
f(x)=arcsin 2x/(1+x) + √(1-2x-3x^2)
case 1:
1-2x-3x^2 > 0
3x^2+2x-1<0
(3x-1)(x+1)<0
-1
case 2:
1+x 不等于0
x不等于-1
and
case 3:
-1<=2x <= 1
-1/2 <=x <=1/2
定义域
case 1 and case 2 and case 3
-1ie -1/2 <=x < 1/3是不是哪里出错了?书上只有个答案是[-1/3,1/3]case 3:-1<=2x <= 1-1/2 <=x <=1/2这个是根据的什么啊？详细点好吗？我以为是(arcsin 2x)/(1+x)现在是arcsin 【2x/(1+x)】case 1:1-2x-3x^2 >= 03x^2+2x-1<0(3x-1)(x+1)<0-1<= x<=1/3case 2:-1<=2x/(1+x)<=1(x-1)/(1+x) <=0and (3x+1)/(1+x) >=01=-1/3}ie -1/3<= x <=1定义域 case 1 and case2-1=
case 1:
1-2x-3x^2 > 0
3x^2+2x-1<0
(3x-1)(x+1)<0
-1
case 2:
1+x 不等于0
x不等于-1
and
case 3:
-1<=2x <= 1
-1/2 <=x <=1/2
定义域
case 1 and case 2 and case 3
-1
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