Question

If $f\left( x \right)=3x-5$, then find ${{f}^{-1}}\left( x \right)$.
A. $\dfrac{1}{3x-5}$
B. $\dfrac{x+5}{3}$
C. does not exist because f is not one-one.
D. does not exist because f is not onto.

Answer 1

Hint: We first need to describe how and why $f\left( x \right)=3x-5$ is a function. We find the characteristics of the function $f\left( x \right)=3x-5$. Then we find the inverse of the function by expressing x as a relation of y. We check if that relation is a function or not.

Complete step by step answer:
We have been given a function in the form of $y=f\left( x \right)=3x-5$.
A function exits only when the function is not one to many. It has to get its fixed image for a particular preimage.
In our given function $f\left( x \right)=3x-5$, the function is one-one function. This means if a and b belong to the domain and $a=b$ then obviously $f\left( a \right)=f\left( b \right)$.
We need to find the function ${{f}^{-1}}\left( x \right)$.
We try to express x with respect to y from the relation $y=f\left( x \right)=3x-5$.
$\begin{align}
  & y=3x-5 \\
 & \Rightarrow 3x=y+5 \\
 & \Rightarrow x=\dfrac{y+5}{3} \\
\end{align}$
Now x is expressed as a relation of y. We need to check if it can be a function or not. We need to check if it can be one to many.
Let’s assume for a fixed value c replaced in place of y, we got two solutions. This means there exists two such x for which we have one value of y.
Then applying this concept for the function $y=f\left( x \right)=3x-5$, the function becomes two to one function. But as we already proved that the function is one-one function, we come to a contradiction.
So, we got $x=\dfrac{y+5}{3}$ as a function. This is the inverse function of $y=f\left( x \right)=3x-5$.
Therefore, $x={{f}^{-1}}\left( y \right)=\dfrac{y+5}{3}$ which gives ${{f}^{-1}}\left( x \right)=\dfrac{x+5}{3}$.

So, the correct answer is “Option B”.

Note: We need to cross-check if a relation is function or not. We have to consider that a function has an inverse only when the function is one-one. This is the condition which the inverse function also has to satisfy. In the end we just need to replace or change the variable to get the required inverse form.