Question

How do you verify that $f\left( x \right) = 3x + 5$, $g\left( x \right) = \dfrac{1}{3}x - \dfrac{5}{3}$ are inverse?

Answer 1

Hint: An inverse function is a function, which can reverse into another function. In other words, if any function ‘f’ takes p to q then the inverse of ‘f’ will take q to p. A function accepts a value followed by performing particular operations on these values to generate an output. If we consider functions, f and g are inverses, then $f\left( {g\left( x \right)} \right)$ is equal to $g\left( {f\left( x \right)} \right)$.
Properties of inverse functions:
 Two functions f and g will be referred to as an inverse of each other if:
Both f and g are one to one functions. In one to one functions, each value is mapped in their domain to exactly one value in the co-domain (range).
The co-domain (range) of one function is the domain of another function and vice versa.

Complete step-by-step solution:
In this question, two functions $f\left( x \right) = 3x + 5$ and $g\left( x \right) = \dfrac{1}{3}x - \dfrac{5}{3}$ are given.
The functions $f\left( x \right) = 3x + 5$ and $g\left( x \right) = \dfrac{1}{3}x - \dfrac{5}{3}$ are inverse if the value of $f\left( {g\left( x \right)} \right)$ is equal to the value of $g\left( {f\left( x \right)} \right)$.
First, let us find the value of $f\left( {g\left( x \right)} \right)$.
$ \Rightarrow f\left( {g\left( x \right)} \right) = 3\left( {\dfrac{1}{3}x - \dfrac{5}{3}} \right) + 5$
Let us take $\dfrac{1}{3}$ as a common factor.
 $ \Rightarrow f\left( {g\left( x \right)} \right) = 3 \times \dfrac{1}{3}\left( {x - 5} \right) + 5$
That is equal to,
$ \Rightarrow f\left( {g\left( x \right)} \right) = x - 5 + 5$
Therefore,
$ \Rightarrow f\left( {g\left( x \right)} \right) = x$
Now, let us find the value of $g\left( {f\left( x \right)} \right)$.
$ \Rightarrow g\left( {f\left( x \right)} \right) = \dfrac{1}{3}\left( {3x + 5} \right) - \dfrac{5}{3}$
Let us multiply$\dfrac{1}{3}$ into the bracket.
$ \Rightarrow g\left( {f\left( x \right)} \right) = \dfrac{1}{3}\left( {3x} \right) + \dfrac{1}{3}\left( 5 \right) - \dfrac{5}{3}$
That is equal to,
$ \Rightarrow g\left( {f\left( x \right)} \right) = x + \dfrac{5}{3} - \dfrac{5}{3}$
Therefore,
$ \Rightarrow g\left( {f\left( x \right)} \right) = x$
Here, the value of $f\left( {g\left( x \right)} \right)$ is equal to x, and the value of $g\left( {f\left( x \right)} \right)$ is also equal to x.
So, we can say that $f\left( {g\left( x \right)} \right) = g\left( {f\left( x \right)} \right)$ .

Hence, $f\left( x \right) = 3x + 5$ and $g\left( x \right) = \dfrac{1}{3}x - \dfrac{5}{3}$ are inverse functions.

Note: There is another method to solve this question.
First, we will take:
$ \Rightarrow f\left( x \right) = y$...(1)
That is equal to,
$ \Rightarrow 3x + 5 = y$
Now, interchange x and y and solve for y.
 So,
$ \Rightarrow f\left( y \right) = x$
Substitute the function value.
$ \Rightarrow 3y + 5 = x$
Let us subtract 5 on both sides.
$ \Rightarrow 3y + 5 - 5 = x - 5$
That is equal to,
$ \Rightarrow 3y = x - 5$
Let us divide both sides by 3.
$ \Rightarrow \dfrac{1}{3}\left( {3y} \right) = \dfrac{1}{3}\left( {x - 5} \right)$
So,
$ \Rightarrow y = \dfrac{1}{3}x - \dfrac{5}{3}$
$ \Rightarrow y = g\left( x \right)$ ...(2)
From equations (1) and (2),
$ \Rightarrow f\left( x \right) = g\left( x \right)$
Hence, $f\left( x \right) = 3x + 5$ and $g\left( x \right) = \dfrac{1}{3}x - \dfrac{5}{3}$ are inverse functions.