Question

Let $f:\left( 0,1 \right)\to \mathbb{R}$ be defined by $f\left( x \right)=\dfrac{b-x}{1-bx}$, where b is constant such that $0< b< 1$, then
A. $f$ is not invertible on $\left( 0,1 \right)$
B. $f\ne {{f}^{-1}}$ on $\left( 0,1 \right)$ and ${{f}^{'}}\left( b \right)=-\dfrac{1}{{{f}^{'}}\left( 0 \right)}$
C. $f={{f}^{-1}}$ on $\left( 0,1 \right)$ and ${{f}^{'}}\left( b \right)=\dfrac{1}{{{f}^{'}}\left( 0 \right)}$
D. ${{f}^{-1}}$ is invertible on $\left( 0,1 \right)$

Answer 1

Hint: We first try to find the inverse function for $f\left( x \right)=\dfrac{b-x}{1-bx}$. The domain for the inverse function would be the same as the range of the main function $f\left( x \right)=\dfrac{b-x}{1-bx}$. Then we try to point out the parts in the domain which are not well defined for the inverse function.

Complete step by step solution:
It is given that $f:\left( 0,1 \right)\to \mathbb{R}$ be defined by $f\left( x \right)=\dfrac{b-x}{1-bx}$.
We first try to find the inverse of the given function.
We take $y=f\left( x \right)=\dfrac{b-x}{1-bx}$.
We try to find the value of $x$ based on $y$.
Therefore,
$\begin{align}
  & y=\dfrac{b-x}{1-bx} \\
 & \Rightarrow y-ybx=b-x \\
 & \Rightarrow x\left( 1-by \right)=b-y \\
 & \Rightarrow x=\dfrac{b-y}{1-by} \\
\end{align}$
Therefore, the inverse of the function $f$ is ${{f}^{-1}}=\dfrac{b-y}{1-by}$.
We have that the domain for $f$ is $\left( 0,1 \right)$ which becomes the range of the function ${{f}^{-1}}$ and the domain of ${{f}^{-1}}$ is $\mathbb{R}$.
Now in ${{f}^{-1}}=\dfrac{b-y}{1-by}$, the denominator can’t be 0.
We can see for the value of $y=\dfrac{1}{b}$, the value of $1-by$ becomes 0.
We have $0< b< 1$ which means $y=\dfrac{1}{b}\in \left( 1,\infty \right)$.
So, in the domain of ${{f}^{-1}}=\dfrac{b-y}{1-by}$, there are parts which cannot define the inverse formula.
Therefore, $f$ is not invertible on $\left( 0,1 \right)$. The correct option is A.

Note: Now we find the differentiation of the $f\left( x \right)=\dfrac{b-x}{1-bx}$.
We get ${{f}^{'}}\left( x \right)=\dfrac{-\left( 1-bx \right)+b\left( b-x \right)}{{{\left( 1-bx \right)}^{2}}}=\dfrac{{{b}^{2}}-1}{{{\left( 1-bx \right)}^{2}}}$. This gives ${{f}^{'}}\left( b \right)=\dfrac{{{b}^{2}}-1}{{{\left( 1-bx \right)}^{2}}}=\dfrac{1}{{{b}^{2}}-1}$
We also have ${{f}^{'}}\left( 0 \right)=\dfrac{{{b}^{2}}-1}{{{\left( 1-bx \right)}^{2}}}={{b}^{2}}-1$
This means ${{f}^{'}}\left( b \right)=\dfrac{1}{{{f}^{'}}\left( 0 \right)}$.