Question

Prove that the function $f:R \to R$ defined by $f\left( x \right) = 4x + 3,$ is invertible and find the inverse of $f$.

Answer 1

Hint: First show that the given function is one-to-one using the concept that:- if \[f\left( {{x_1}} \right) = f\left( {{x_2}} \right) \Rightarrow {x_1} = {x_2}\] then the function is one-to-one. After that show that the function is onto. Once the function is one to one and onto then it concludes that the function is invertible.

Complete step by step solution:
We have given a function $f:R \to R$ defined by $f\left( x \right) = 4x + 3$.
The goal is to show that this given function is invertible and then find the inverse of the function.
First of all, if we want to show that the given function is invertible, we need to show that the given function is both one-to-one and onto.
First show that the function is one-to-one. So, assume that:
$f\left( {{x_1}} \right) = 4{x_1} + 3$ and $f\left( {{x_2}} \right) = 4{x_2} + 3$
We can say that the function is one-to-one, if
\[f\left( {{x_1}} \right) = f\left( {{x_2}} \right) \Rightarrow {x_1} = {x_2}\]
So, take \[f\left( {{x_1}} \right) = f\left( {{x_2}} \right)\] and substitute the values,
$4{x_1} + 3 = 4{x_2} + 3$
$ \Rightarrow 4{x_1} = 4{x_2}$
$ \Rightarrow {x_1} = {x_2}$
We have seen that \[f\left( {{x_1}} \right) = f\left( {{x_2}} \right) \Rightarrow {x_1} = {x_2}\], thus the function is one-to-one.
Now, show that the function is onto.
We have to show that for every $y \in R$, there exists $x \in R$, such that $f\left( x \right) = y$.
Now let $f\left( x \right) = y$, for some $y \in R$, then
$4x + 3 = y$
Now, solve the equation for $x$,
$4x = y - 3$
$x = \dfrac{{y - 3}}{4}$
Now, check for $y = f\left( x \right)$by substituting the value of $x$ into the function.
\[f\left( x \right) = f\left( {\dfrac{{y - 3}}{4}} \right)\]
\[f\left( x \right) = 4\left( {\dfrac{{y - 3}}{4}} \right) + 3\]
\[f\left( x \right) = \left( {y - 3} \right) + 3\]
\[f\left( x \right) = y\]
Thus, for every $y \in R$, there exists some $x \in R$ such that $f\left( x \right) = y$. Thus the function $f\left( x \right)$ is onto.
As we had shown that the function is both one-to-one and onto, it means that the function is invertible.

Thus, the function $f\left( x \right) = 4x + 3$ is invertible.

Note: While showing that the function is onto we have to take care about the domain and range of the function. The main concept is to show that for all values of the range of the function, there exists a value in the domain such that the function is defined.