Question

Consider the function $f\left( x \right)=\dfrac{x-2}{x-3}$ defined by f: A$\to $B where $A=R-\left\{ 3 \right\}$,$B=R-\left\{ 1 \right\}$. Show that f(x) is one-one and onto and hence find ${{f}^{-1}}$.

Answer 1

Hint: To prove that the function f(x) is one-one and onto we should know the method of proof adopted for each of the conditions. For a function to be one-one function, it should satisfy the condition that if $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ implies that ${{x}_{1}}={{x}_{2}}$. For a function to be onto, we should take $f\left( x \right)=y$ and get the value of x in terms of y and back substitute the value of x in the function f(x). If we get y in the back calculation without any undetermined point, then the function is onto function. Once we know the function is both one-one and onto, we can call it bijective and we can conclude that the function has an inverse function. We can calculate the inverse function by assuming${{f}^{-1}}\left( x \right)=y\Rightarrow x=f\left( y \right)$, and get the value of y in terms of x. The final value of y in terms of x will be the inverse function of f(x).

Complete step by step answer:
In the question, we are given a function $f\left( x \right)=\dfrac{x-2}{x-3}$.
Let us consider the one-one property.
For a function f(x) to be one-one function, it should satisfy the condition that if $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ implies that ${{x}_{1}}={{x}_{2}}$.
Let us consider two values ${{x}_{1}},{{x}_{2}}$ for which $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$. We can write that
$\dfrac{{{x}_{1}}-2}{{{x}_{1}}-3}=\dfrac{{{x}_{2}}-2}{{{x}_{2}}-3}$
By cross multiplying, we get
$\begin{align}
  & \left( {{x}_{1}}-2 \right)\times \left( {{x}_{2}}-3 \right)=\left( {{x}_{2}}-2 \right)\times \left( {{x}_{1}}-3 \right) \\
 & {{x}_{1}}{{x}_{2}}-2{{x}_{2}}-3{{x}_{1}}+6={{x}_{1}}{{x}_{2}}-2{{x}_{1}}-3{{x}_{2}}+6 \\
\end{align}$
Cancelling the terms, we get
$\begin{align}
  & -{{x}_{1}}=-{{x}_{2}} \\
 & {{x}_{1}}={{x}_{2}} \\
\end{align}$
So, we got that $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\Rightarrow {{x}_{1}}={{x}_{2}}$
So, f(x) is a one-one function.
Let us consider $f\left( x \right)=y$, we can write that
$\begin{align}
  & y=\dfrac{x-2}{x-3} \\
 & xy-3y=x-2 \\
 & xy-x=3y-2 \\
 & x\left( y-1 \right)=3y-2 \\
 & x=\dfrac{3y-2}{y-1} \\
\end{align}$
Let us substitute the value of x back in $f\left( x \right)=\dfrac{x-2}{x-3}$, we get
$f\left( x \right)=\dfrac{\dfrac{3y-2}{y-1}-2}{\dfrac{3y-2}{y-1}-3}=\dfrac{\dfrac{3y-2-2y+2}{y-1}}{\dfrac{3y-2-3y+3}{y-1}}=\dfrac{y}{1}=y$.
In the above calculations, we have cancelled (y-1) because the range of y-1 does not include 1. If the range of f(x) also includes the value of 1, we cannot cancel y-1 and the function f(x) will not be onto.
In our question, function f(x) is one-one and onto function in the given range.
We know that for a one-one and onto function, inverse exists.
Let us consider ${{f}^{-1}}\left( x \right)=y$, we can write that
$\begin{align}
  & x=f\left( y \right) \\
 & x=\dfrac{y-2}{y-3} \\
 & xy-3x=y-2 \\
 & xy-y=3x-2 \\
 & y\left( x-1 \right)=3x-2 \\
 & y=\dfrac{3x-2}{x-1} \\
\end{align}$
But we know that ${{f}^{-1}}\left( x \right)=y$, we can write
${{f}^{-1}}\left( x \right)=\dfrac{3x-2}{x-1}$
$\therefore $It is proved that the function $f\left( x \right)=\dfrac{x-2}{x-3}$ is one-one and onto in the given domain and range and ${{f}^{-1}}\left( x \right)=\dfrac{3x-2}{x-1}$.

Note:
The domain and range in the question play a major role in deciding if the function is one-one or not and onto or not. In our question, at x=3 there is no value for the function, hence we can infer that 3 is removed from the domain. Similarly, for f(x)=1, we don’t have a value of x, that is why it is removed from the range to make the function one-one and onto function. We can also see that the end of the proof for onto function has the inverse of the function.