Proving the invertibility of a real-valued function 
Let $f:(0,1) \to \mathbb R$ be defined by $f(x)= (b-x)/(1-bx)$ , where $b$ is a constant such that $0<b<1$. Is $f$ invertible on $(0,1)$?

I solved the above question via two methods.


*

*Through composition criteria.
i.e.  if  $g[f(x)]= x$ and $f[g(x)]=y$ holds, then $f$ is invertible.

*If a function is one-to-one and onto, it is invertible.
The function comes out to be invertible by the first method.
But is not onto.
Can some one help me figure out the mistake?
 A: The denominator $1-bx$ vanishes for $x=1/b>1$, so the function is indeed defined on $(0,1)$. Now let's solve
$$
y=\frac{b-x}{1-bx}
$$
with respect to $x$, getting
$$
x=\frac{b-y}{1-by}
$$
Note also that
$$
\frac{b-x}{1-bx}=\frac{1}{b}
$$
has no solution, so $y$ never gets the value $1/b$. Therefore the function is invertible.
How can we determine the domain of the inverse? Since $f$ is invertible and obviously continuous, it is either strictly increasing or strictly decreasing. Hence we can compute the limits:
$$
\lim_{x\to0^+}f(x)=b,
\qquad
\lim_{x\to1^-}f(x)=-1
$$
Thus the range of $f$ (and the domain of the inverse) is $(-1,b)$.
A: If $b\in (0,1)$ then $f(x)=\frac{b-x}{1-bx}$ is defined  and strictly decreasing in $[-1,1]$. Moreover it is easy to see that $f([-1,1])=[-1,1]$. Hence $f$ is bijective from $[-1,1]$ onto $[-1,1]$. Since $f(f(x))=x$ then the inverse is $f$ itself.
It follows that $f$ restricted to $(0,1)$ is invertible (with respect to its image $f((0,1))=(-1,b)$) and its inverse is $f$ restricted to $(-1,b)$.
