Question

Show that the function $f\left( x \right)=\dfrac{4x+3}{6x-4}$ defined in $A=R-\left\{ \dfrac{2}{3} \right\}$. 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 is 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 solution:
In the question, we are given a function$f\left( x \right)=\dfrac{4x+3}{6x-4}$.
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
$\dfrac{4{{x}_{1}}+3}{6{{x}_{1}}-4}=\dfrac{4{{x}_{2}}+3}{6{{x}_{2}}-4}$
By cross multiplying, we get
$\begin{align}
  & \left( 4{{x}_{1}}+3 \right)\times \left( 6{{x}_{2}}-4 \right)=\left( 4{{x}_{2}}+3 \right)\times \left( 6{{x}_{1}}-4 \right) \\
 & 24{{x}_{1}}{{x}_{2}}+18{{x}_{2}}-16{{x}_{1}}-12=24{{x}_{1}}{{x}_{2}}+18{{x}_{1}}-16{{x}_{2}}-12 \\
\end{align}$
Cancelling the terms, we get
$\begin{align}
  & 34{{x}_{1}}=34{{x}_{2}} \\
 & {{x}_{1}}={{x}_{2}} \\
\end{align}$
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
$\begin{align}
  & y=\dfrac{4x+3}{6x-4} \\
 & 6xy-4y=4x+3 \\
 & 6xy-4x=4y+3 \\
 & x\left( 6y-4 \right)=4y+3 \\
 & x=\dfrac{4y+3}{6y-4} \\
\end{align}$
Let us substitute the value of x back in$f\left( x \right)=\dfrac{4x+3}{6x-4}$, we get
$f\left( x \right)=\dfrac{4\times \dfrac{4y+3}{6y-4}+3}{6\times \dfrac{4y+3}{6y-4}-4}=\dfrac{\dfrac{16y+12+18y-12}{6y-4}}{\dfrac{24y+18-24y+16}{6y-4}}=\dfrac{34y}{34}=y$.
In the above calculations, we have canceled (6y-4) because the range of 6y-4 does not include $\dfrac{2}{3}$. If the range of f(x) also includes the value of $\dfrac{2}{3}$, we cannot cancel 6y-4 and the function f(x) will not be onto function.
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
$\begin{align}
  & x=f\left( y \right) \\
 & x=\dfrac{4y+3}{6y-4} \\
 & 6xy-4x=4y+3 \\
 & 6xy-4y=4x+3 \\
 & y\left( 6x-4 \right)=4x+3 \\
 & y=\dfrac{4x+3}{6x-4} \\
\end{align}$
But we know that ${{f}^{-1}}\left( x \right)=y$, we can write
${{f}^{-1}}\left( x \right)=\dfrac{4x+3}{6x-4}$
$\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 ${{f}^{-1}}\left( x \right)=\dfrac{4x+3}{6x-4}$

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 f(x)=$\dfrac{2}{3}$, 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. In our question, we can infer that the function f(x) is equal to its inverse. In these type of special functions, we can write a property
$\begin{align}
  & f(x)={{f}^{-1}}\left( x \right) \\
 & f\left( f\left( x \right) \right)=x \\
\end{align}$.
This is a useful property in some type of question.