Question

求助大一高数题

求函数定义域y=arcsin((x-1)/3)
求函数的反函数y=(2^x)/(1+2^x)

Answer 1

1） -1 <= (x-1)/3 <= 1, 解得 -2 <= x <= 4
2) 由已知2^x = y/(1-y), 即x = log2(y/1-y), 即反函数为y = log2(x/1-x), x∈(0, 1)

Answer 2

解1 (x-1)/3的取值范围是[-1,1],故x的范围[-2,4]
2 2^x = y/(1-y), 即x = log2(y/1-y), 即反函数为y = log2(x/1-x), x∈(0, 1)

Answer 3

(1)-1≤(x-1)/3≤1,
-3≤x-1≤3
-2≤x≤4
(2)y+2^x y=2^x
2^x=y/(1-y)
x=log(2)(y/(1-y))
反函数y=log(2)(x/(1-x))
定义域 0<x<1

Answer 4

1解：-1<=（x-1）/3<=1
求得：-2<=x<=4
2解：1/y=(1+2^x)/(2^x)
1/y=1/(2^x) + 1
1/y - 1=1/(2^x)
y/(1-y)=2^x
x=log(2)(y/(1-y))

Answer 5

-2<X<4