Question

这些反函数怎么求？第3、5、6题！

希望能帮解答，谢谢！

Answer 1

1. y^3=x+1, x=y^3-1, 所以 y=x^3-1
2.y+xy=1-x, x(y+1)=1-y, x=(1-y)/(1+y), 所以 y=(1-x)/(1+x)
3.cxy+dy=ax+b, x(cy-a)=b-dy, x=(b-dy)/(cy-a), 所以 y=(b-dx)/(cx-a)
4. -π/6<=x<=π/6, -π/2<=3x<=π/2, -1<=sin3x<歼好=1, -2<派扒=y<=2
six3x=y/2, 3x=arcsin(y/2), x=1/3arcsin(y/尘改昌2), 所以 y=1/3arcsin(x/2) (-2<=x<=2)
5.y-1=ln(x+2), x+2= e^(y-1), x=e^(y-1)-2, 所以 y=e^(x-1)-2
6.2^xy+y=2^x, 2^x(1-y)=y, 2^x=y/(1-y), x=log2[y/(1-y)], 所以 y=log2[x/(1-x)]

Answer 2

(3)cxy+dy=ax+b
(cy-a)x=b-dy
x=(b-dy)/慧扰(cy-a)
交换x，y，得
反函数为
y=(b-dx)/(cx-a)，x≠a/c
(5) y-1=ln(x+2)
x+2=e^(y-1)
x=e^(y-1) -2
交换x，y，得
反函数为y=e^(x-1) -2，罩纳x∈R
(6) y·2^x+y=2^x
(1-y)·2^x=y
2^x=y/(1-y)
x=log₂[y/(1-y)]
交换物碧没x，y，得
反函数为
y=log₂[x/(1-x)]，0<x<1

Answer 3

答：
反纯李函数即原函做盯迟数的x和y互换。
(3).
x=(ay+b)/(cy+d)
cxy+dx=ay+b
(cx-a)y=b-dx
y=(b-dx)/(cx-a)
(5)
x=1+ln(y+2)
x-1=ln(y+2)
e^(x-1)=y+2
y=e^(x-1)-2
(6)
x=2^y/(2^y+1)
x=(2^y+1-1)/(2^y+1)
x=1-1/(2^y+1)
1/(2^y+1)=1-x
2^y+1=1/(1-x)
2^y=1/(1-x)-1
2^y=x/(1-x)
y=log(2) (x/则运(1-x))

Answer 4

（3）
(cx+d)y=ax+b
cxy+dy=ax+b
cxy-ax=b-dy
x(cy-a)=b-dy
x=(b-dy)/(cy-a) 让x和y互换
得到y=(b-dx)/(cx-a)

(5)
y=l+ln(x+2)
y-1=ln(x+2)

e^(x+2)=y-1
e^(y-1)=x+2

x=e^(y-1)-2 让x和y互换
y=e^(x-1)-2

(6)

令2^x=u
则方程变为
y=u/(u+1)
y=1-(1/(u+1))
u=[1/(1-y)]-1
u=y/(1-y)
2^x=y/(1-y)

x=log(2) y/(1-y) 让x和y互换

y=log(2) y/(1-x) (2)的意唤缺思是2为底的对轿运数闭链梁

Answer 5