Question

Let $f:R \to R$ be defined as $f\left( x \right) = 10x + 7$ then find the function $g:R \to R$ such that $gof = fog = {I_R}$.

Answer 1

Hint: In this problem, first we will prove that the given function $f\left( x \right) = 10x + 7$is invertible. Then, we will use the given condition $gof = fog = {I_R}$ to find the required function $g\left( x \right)$. That is, we will find the inverse function of $f\left( x \right)$.

Complete step-by-step answer:
In this problem, the given function is $f:R \to R$, $f\left( x \right) = 10x + 7$. We need to find the function $g:R \to R$ such that $gof = fog = {I_R}$. Note that $gof = fog = {I_R}$ implies $g\left( x \right)$ is an inverse function of $f\left( x \right)$. That is, $g\left( x \right) = {f^{ - 1}}\left( x \right)$. Also we can write $f\left( x \right) = {g^{ - 1}}\left( x \right)$. To find the required function $g\left( x \right)$, we will find the inverse of $f\left( x \right)$. For this, first we will prove that the function $f\left( x \right)$ is invertible. That is, we will prove that the function is one-one and onto both. Note that here the domain of $f$ is $R$. Let us take two elements ${x_1}$ and ${x_2}$ from the domain $R$. Let us consider $f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$. So, we can write
 $f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$
$ \Rightarrow 10{x_1} + 7 = 10{x_2} + 7$
$ \Rightarrow 10{x_1} = 10{x_2}$
$ \Rightarrow {x_1} = {x_2}$
Here we get $f\left( {{x_1}} \right) = f\left( {{x_2}} \right) \Rightarrow {x_1} = {x_2}$. So, we can say that $f$ is a one-one function.
Let us take an arbitrary element $y$ from codomain $R$. Let us consider $y = f\left( x \right)$. So, we can write
$y = f\left( x \right)$
$ \Rightarrow y = 10x + 7$
$ \Rightarrow y - 7 = 10x$
$ \Rightarrow x = \dfrac{{y - 7}}{{10}}$
We can see that $x = \dfrac{{y - 7}}{{10}} \in R$(domain). That is, for arbitrary element $y \in R$ (codomain) there is an element $x \in R$ (domain) such that $f\left( x \right) = y$. So, we can say that $f$ is onto function.
Here $f:R \to R$, $f\left( x \right) = 10x + 7$ is one-one and onto. So, we can say that $f$ is invertible. That is, ${f^{ - 1}}\left( x \right)$ exists. To find ${f^{ - 1}}\left( x \right)$, let us consider $y = f\left( x \right)$. So, we can write
$y = 10x + 7$
$ \Rightarrow x = \dfrac{{y - 7}}{{10}}$
$ \Rightarrow {f^{ - 1}}\left( y \right) = \dfrac{{y - 7}}{{10}}\quad \left[ {\because y = f\left( x \right) \Rightarrow x = {f^{ - 1}}\left( y \right)} \right]$
Replace $y$ by $x$, we get
${f^{ - 1}}\left( x \right) = \dfrac{{x - 7}}{{10}}$
$ \Rightarrow g\left( x \right) = \dfrac{{x - 7}}{{10}}\quad \left[ {\because gof = fog = {I_R} \Rightarrow g\left( x \right) = {f^{ - 1}}\left( x \right)} \right]$
Hence, the required function is $g:R \to R$, $g\left( x \right) = \dfrac{{x - 7}}{{10}}$.

Note: If $f:A \to B$ and $g:B \to c$ are two functions then the composition of $f$ and $g$ is denoted by $gof$ and it is defined as $gof:A \to c$, $gof\left( x \right) = g\left[ {f\left( x \right)} \right]$, $\forall x \in A$. If the function $f$ is invertible then $f$ must be one-one and onto. Converse is also true.