求反函数y=1+sin(x-1/x+1) y=(sin3x)/2 y=2^x/(2^x+1)

1.y=1+sin(x-1/x+1) ,
(x-1)/(x+1)=kπ+(-1)^k*arcsin(y-1),k∈Z,
x[1-kπ-(-1)^k*arcsin(y-1)]=1+kπ+(-1)^k*arcsin(y-1),
x=[1+kπ+(-1)^k*arcsin(y-1)]/[1-kπ-(-1)^k*arcsin(y-1)],
x,y互换得y=[1+kπ+(-1)^k*arcsin(x-1)]/[1-kπ-(-1)^k*arcsin(x-1)].
2.y=(sin3x)/2
仿上,sin3x=2y,
3x=kπ+(-1)^k*arcsin(2y),
x=[kπ+(-1)^k*arcsin(2y)]/3,
y=[kπ+(-1)^k*arcsin(2x)]/3.
3.y=2^x/(2^x+1),
y*2^x+y=2^x,
2^x=y/(1-y),
x=log[y/(1-y)],
y=log[x/(1-x)].