Question

If $f\left( x \right)=\dfrac{4x+3}{6x-4},x\ne \dfrac{2}{3}$ , show that $f\circ f\left( x \right)=x$ , for all $x\ne \dfrac{2}{3}$ . What is the inverse of f?

Answer 1

Hint: To show that $f\circ f\left( x \right)=x$ , we have to write the definition of the composite function mathematically, that is, $f\circ f\left( x \right)$ is found by substituting x as $f\left( x \right)$ in f(x). To find the inverse of a function, firstly, we have to replace y with f(x). Then, we have to replace y with x and solve for y in terms of x. Finally, we have to replace y with ${{f}^{-1}}\left( x \right)$ .

Complete step by step answer:
We are given that $f\left( x \right)=\dfrac{4x+3}{6x-4}$ . We have to show that $f\circ f\left( x \right)=x$ . We know that the function composition with the same function, that is $f\circ f\left( x \right)$ is found by substituting x as $f\left( x \right)$ in f(x).
$\begin{align}
  & \Rightarrow f\circ f\left( x \right)=f\left( f\left( x \right) \right) \\
 & \Rightarrow f\circ f\left( x \right)=\dfrac{4\left( \dfrac{4x+3}{6x-4} \right)+3}{6\left( \dfrac{4x+3}{6x-4} \right)-4} \\
\end{align}$
Let us apply the distributive property.
\[\begin{align}
  & \Rightarrow f\circ f\left( x \right)=\dfrac{\dfrac{16x+12+3\left( 6x-4 \right)}{\require{cancel}\cancel{6x-4}}}{\dfrac{24x+18-4\left( 6x-4 \right)}{\require{cancel}\cancel{6x-4}}} \\
 & \Rightarrow f\circ f\left( x \right)=\dfrac{16x+12+3\left( 6x-4 \right)}{24x+18-4\left( 6x-4 \right)} \\
\end{align}\]
We have to apply distributive property.
\[\begin{align}
  & \Rightarrow f\circ f\left( x \right)=\dfrac{16x+\require{cancel}\cancel{12}+18x-\require{cancel}\cancel{12}}{\require{cancel}\cancel{24x}+18-\require{cancel}\cancel{24x}+16} \\
 & \Rightarrow f\circ f\left( x \right)=\dfrac{16x+18x}{18+16} \\
 & \Rightarrow f\circ f\left( x \right)=\dfrac{\require{cancel}\cancel{34}x}{\require{cancel}\cancel{34}} \\
 & \Rightarrow f\circ f\left( x \right)=x \\
\end{align}\]
Hence, proved.
Now, we have to find the inverse of $f\left( x \right)=\dfrac{4x+3}{6x-4}$ . Firstly, we have to replace $f\left( x \right)$ as y.
$\Rightarrow y=\dfrac{4x+3}{6x-4}$
Now, we have to replace x by y.
$\Rightarrow x=\dfrac{4y+3}{6y-4}$
We have to solve for y in terms of x. For this, we will take the denominator in the RHS to the LHS.
$\Rightarrow x\left( 6y-4 \right)=4y+3$
Let us apply distributive property on the LHS.
$\Rightarrow 6xy-4x=4y+3$
Now, we have to move 4y to the LHS and 4x to the RHS.
$\Rightarrow 6xy-4y=4x+3$
Let us take the common term from the LHS outside.
$\Rightarrow y\left( 6x-4 \right)=4x+3$
Now, we have to move the coefficient of y to the RHS.
$\Rightarrow y=\dfrac{4x+3}{6x-4}$
Finally, we have to replace y with ${{f}^{-1}}\left( x \right)$ .
$\Rightarrow {{f}^{-1}}\left( x \right)=\dfrac{4x+3}{6x-4}$
Therefore, the inverse of f(x) is $\dfrac{4x+3}{6x-4}$ .

Note: Students must know the concept of function composition and its properties. They must be thorough with the steps to find the inverse of a function. They must know how to solve algebraic equations and the rules associated with it.