How do I prove that $f(x)$ is not invertible in $(0,1)$? Let $f:(0,1)\to \mathbb R$ be defined by $f(x)=\frac{b−x}{1−bx}$, where $b$ is a constant such that $0<b<1$.
How do I prove that $f(x)$ is not invertible in $(0,1)$?
P.S:I was able to prove that $f(x)$ is one-one...
 A: Let $a\in\Bbb R$ and suppose $f(x)=a$.
Then $\frac{b-x}{1-bx}=a$.
Solving for $x$ we have $x=\frac{b-a}{1-ba}$
Thus if $a=\frac1b$ then there can be no such $x$ that maps to it.
Thus $f$ is not onto.  Thus $f$ is not invertible.
A: We have $f: (0,1) \to \mathbb{R}, x \mapsto \frac{b-x}{1-bx}$.
Now, we note that $f$ is continuous and injective, so in particular we must have that $f$ is increasing or decreasing.
Then we have $\hat{f}(0) = \lim_{x \to 0} f(x)$ is well-defined on the extended reals. In case $f$ is invertible, we must have $\hat{f}(0) = \pm \infty$, else the range of $f$ is bounded below or above, so in particular not the whole of $\mathbb{R}$, we have
$$ \hat{f}(0) = \lim_{x \to 0} \frac{b-x}{1-bx} = \frac{b}{1} = b$$
as the both parts of the fraction are continuous and the denominator is bounded away from zero.
As $0 < b <1$, we have that $f$ is not-invertible when defined from $(0,1) \to \mathbb{R}$.
A: $f(x)=b\iff b-x=b-b^2x\iff -x=-b^2x\iff x=0.$
A: $f'(x) = (-(1-bx)+b(b-x))/(1-bx)^2 = (b^2-1)/(1-bx)^2<0$ since $1<b$ thus $f$ decreases strictly and $f:(0,1)\rightarrow f(0,1)$ is bijective. But remak that $f(0,1)$ is not contained in $(0,1)$ since $f(1)=-1$. So $f$ is bijective onto its image.
