Question

How do find the inverse of \[y=\dfrac{1}{x-3}\]?

Answer 1

Hint: In this question, we have to find the inverse of a function. This question is from the topic of pre-calculus. During solving this question, firstly we will see what are the steps to find the inverse of a function. After that, we will solve the question. The inverse of a function is the image of the function whose reflecting surface is the line y=x.

Complete answer:
Let us solve this question.
In this question, it is asked to find the inverse of \[y=\dfrac{1}{x-3}\].
So, first let us know how to find the inverse of a function.
If a function y=f(x) is given and we have to find the inverse of that function where f(x) is a function of x.
Then, the first step will be replacing y by x and replacing x by y. This is done to make the rest processes easier.
After that, we will find the value of y in terms of x.
The value of y will be the inverse of function of x.
Using the above methods, we will solve the question.
As \[y=\dfrac{1}{x-3}\], then we will replace y by x and x by y. We get,
\[x=\dfrac{1}{y-3}\]
Now, we will find the value of y with respect to x or in terms of x.
\[\Rightarrow \dfrac{x}{1}=\dfrac{1}{y-3}\]
Doing cross multiplication, we can write the above equation as
\[\Rightarrow \dfrac{y-3}{1}=\dfrac{1}{x}\]
\[\Rightarrow y-3=\dfrac{1}{x}\]
Carrying 3 to the right side of the equation, we get
\[\Rightarrow y=\dfrac{1}{x}+3\]
Taking LCM and solving in the right side of the equation, we get
\[\Rightarrow y=\dfrac{3x+1}{x}\]
Hence, we get that the inverse of \[y=\dfrac{1}{x-3}\] is \[\dfrac{3x+1}{x}\].

Note: We should have a better knowledge in pre-calculus for solving this type of question. While finding the value of y in terms of x after replacing y by x and x by y, mistakes can be made. So, be aware of that. We have a method for verifying that if we are correct or not. Let us suppose we have a function \[f\] which is a function of x. And we have found that the inverse of \[f\] is \[{{f}^{-1}}\]. Then, the relation \[{{f}^{-1}}\left( f\left( x \right) \right)=f\left( {{f}^{-1}}\left( x \right) \right)=x\] will always satisfy. For example, we can see in the equation \[y=\dfrac{1}{x-3}\]. We know that its inverse is \[\dfrac{3x+1}{x}\]. Let us verify that.
Here, function is \[f\left( x \right)=\dfrac{1}{x-3}\] and inverse of the function is \[{{f}^{-1}}\left( x \right)=\dfrac{3x+1}{x}\].
So, \[{{f}^{-1}}\left( f\left( x \right) \right)\] will be \[{{f}^{-1}}\left( f\left( x \right) \right)=\dfrac{3f\left( x \right)+1}{f\left( x \right)}=\dfrac{3\left( \dfrac{1}{x-3} \right)+1}{\left( \dfrac{1}{x-3} \right)}=\dfrac{\dfrac{3+x-3}{x-3}}{\dfrac{1}{x-3}}=\dfrac{\dfrac{x}{x-3}}{\dfrac{1}{x-3}}\] which is equal to x. Hence, it is verified.