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

Last updated date: 20th Jun 2024
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Hint: Before attempting this question one should have prior knowledge about the concept of functions and also remember that $fof\left( x \right) = x$ means $f\left( {f\left( x \right)} \right) = x$ so to proof this use $f\left( {f\left( x \right)} \right) = f\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right)$, use this information to approach the solution.

Complete step-by-step solution:
According to the given information we have function $f\left( x \right) = \dfrac{{4x + 3}}{{6x - 4}},x \ne \dfrac{2}{3}$
First of all, we have to show $fof\left( x \right) = x$ which also means $f\left( {f\left( x \right)} \right) = x$
So, $f\left( {f\left( x \right)} \right) = f\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right)$
Now, substituting the value of $f\left( x \right)$i.e. $\dfrac{{4x + 3}}{{6x - 4}}$
$f\left( {f\left( x \right)} \right) = f\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right) = \dfrac{{4\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right) + 3}}{{6\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right) - 4}}$
$ \Rightarrow $$f\left( {f\left( x \right)} \right) = \dfrac{{\dfrac{{16x + 12 + 18x - 12}}{{6x - 4}}}}{{\dfrac{{24x + 18 - 24x + 16}}{{6x - 4}}}}$
$ \Rightarrow $$f\left( {f\left( x \right)} \right) = \dfrac{{16x + 12 + 18x - 12}}{{24x + 18 - 24x + 16}}$
$ \Rightarrow $$f\left( {f\left( x \right)} \right) = \dfrac{{34x}}{{34}} = x$
Hence proved, $f\left( {f\left( x \right)} \right) = x$
Let $g$be the inverse of the given function
So, we know that if g is inverse of function $f\left( x \right)$
Then $gof\left( x \right) = x$ and $fog\left( x \right) = x$
Now comparing the above statement with $fof\left( x \right) = x$
So, as we can say that after comparing $gof\left( x \right) = x$ with $fof\left( x \right) = x$
Here g is $f$
And on comparing $fog\left( x \right) = x$ with $fof\left( x \right) = x$ again here $f\left( x \right)$ is $g\left( x \right)$
Therefore, we can say that the given function is inverse of itself
Which means $f\left( x \right) = {f^{ - 1}}\left( x \right)$
You can easily see inverse of $f$ is equal to $f\left( x \right)$
Hence, ${f^{ - 1}}\left( x \right) = \dfrac{{4x + 3}}{{6x - 4}}$

Note: In the above solution we came across the term “function” which can be explained as a relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y then the function is represented by $f:X \to Y$ examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.