Question

If we have a function $f(x) = \dfrac{{(4x + 3)}}{{(6x - 4)}}$, $x \ne \dfrac{2}{3}$ , show that $fof(x) = x$ , for all $x \ne \dfrac{2}{3}$ . What is the inverse of $f$ ?

Answer 1

If we substitute $x = \dfrac{2}{3}$ in the given function $f(x) = \dfrac{{(4x + 3)}}{{(6x - 4)}}$ then the value of $f(\dfrac{2}{3})$ will be infinite i.e., we can’t define the function at that value. In order to prove that $fof(x) = x$ first let us consider the L.H.S (i.e., Left Hand Side) of the equation and on solving that equation we get the result which is R.H.S (Right Hand Side) of the equation. Inverse of a function can be calculated by solving the function for $x$ .

Complete step-by-step solution:
Given that $f(x) = \dfrac{{(4x + 3)}}{{(6x - 4)}}$ , for all $x \ne \dfrac{2}{3}$ . In the question it is required to prove $fof(x) = x$ . In order to prove that let us consider the Left Hand Side (L.H.S) of the equation.
Here, $fof(x)$ is a function of the given function i.e., $f(f(x))$ and we can solve this by substituting the function within the function.
${\text{L}}{\text{.H}}{\text{.S of the equation = }}fof(x)$
$= f(f(x))$
Substitute $f(x) = \dfrac{{(4x + 3)}}{{(6x - 4)}}$ in the above function
$= f(\dfrac{{(4x + 3)}}{{(6x - 4)}})$
$= \dfrac{{4 \times (\dfrac{{(4x + 3)}}{{(6x - 4)}}) + 3}}{{6 \times (\dfrac{{(4x + 3)}}{{6x - 4)}}) - 4}}$
On simplification of the above we get the following
$= \dfrac{{4 \times (4x + 3) + 3 \times (6x - 4)}}{{6 \times (4x + 3) - 4 \times (6x - 4)}}$
On further simplification we get the following
$= \dfrac{{16x + 12 + 18x - 12}}{{24x + 18 - 24x + 16}}$
$= \dfrac{{34x}}{{34}}$
$= x$
$= R.H.S$
Therefore, $L.H.S = R.H.S$ $\Rightarrow fof(x) = x$ , for all $x \ne \dfrac{2}{3}$
Inverse of the function $f(x)$ can be obtained by solving the given equation $f(x) = \dfrac{{(4x + 3)}}{{(6x - 4)}}$ for $x$ .
Let $f(x) = y \Rightarrow x = {f^{ - 1}}(y)$ then $f(x) = y = \dfrac{{(4x + 3)}}{{(6x - 4)}}$
Where $f(x)$ is the given function and ${f^{ - 1}}(x)$ is the inverse of the given function.
$y(6x - 4) = (4x + 3)$
$\Rightarrow 6xy - 4y = 4x + 3$
$\Rightarrow 6xy - 4x - 4y - 3 = 0$
On rearranging the above equation we get
$\Rightarrow x(6y - 4) - (4y + 3) = 0$
$\Rightarrow x(6y - 4) = (4y + 3)$
$\Rightarrow x = \dfrac{{(4y + 3)}}{{(6y - 4)}}$
We know that $y = {f^{ - 1}}(x)$
${f^{ - 1}}(y) = \dfrac{{(4y + 3)}}{{(6y - 4)}}$
Therefore, ${f^{ - 1}}(x) = \dfrac{{(4x + 3)}}{{(6x - 4)}} = f(x)$ , for all $x \ne \dfrac{2}{3}$
$\Rightarrow {f^{ - 1}}(x) = f(x)$
${\text{Inverse of the function }}f(x) = f(x)$.

Note: In order to find the values of $x$ at which the function $f(x)$ is not defined we need to equate the denominator of the function to zero. We can also solve the above problem in the following way. After proving $fof(x) = x$ . Now we know that $fof(x) = x \Rightarrow f(x) = {f^{ - 1}}(x)$ i.e., ${f^{ - 1}}(x) = \dfrac{{(4x + 3)}}{{(6x - 4)}}$ . If the question is to find the Inverse of the given function $f(x)$ without providing $fof(x) = x$ , then we can follow the process done in the above solution.