Question

How do you find the inverse of $f(x) = \sqrt {x - 2} $ and is it a function?

Answer 1

Hint: We will first put y = f (x) and then find the value of x in terms of y. now we will do the required replacement and thus get the inverse. Then, we will see if it follows the rules of being a function or not.

Complete step-by-step answer:
We are given that we have $f(x) = \sqrt {x - 2} $ and we need to find its inverse.
Let us put y = f (x) in the given function to obtain the following:-
$ \Rightarrow y = \sqrt {x - 2} $
Squaring both sides of the above equation, we will then obtain:-
$ \Rightarrow {y^2} = x - 2$
Re – arranging the terms and writing the above equation as follows:-
$ \Rightarrow x - 2 = {y^2}$
Taking 2 from subtraction in the left hand side to addition in the right hand side to obtain the following expression:-
$ \Rightarrow x = {y^2} + 2$
Now, as we assumed that y = f (x), we will then obtain: $x = {f^{ - 1}}(y)$. Putting this in the expression and equation we obtained to obtain the following expression:-
$ \Rightarrow {f^{ - 1}}(y) = {y^2} + 2$
Now, replacing y by x in the above equation to obtain:-
$ \Rightarrow {f^{ - 1}}(x) = {x^2} + 2$
Now, since in the function we have a square of x.
So, if we put in x = 2 or x = -2, we still get the same answer because it is being squared.
So, it will not be a function when we take the domain to be real number because one point cannot have multiple images. Therefore, if we take the domain to be greater than or equal to 0, then we can claim that it is a function.

Hence, the inverse of $f(x) = \sqrt {x - 2} $ is ${f^{ - 1}}(x) = {x^2} + 2$ and it is a function when domain is greater than equal to 0.

Note:
The students must notice that domain of the original function f (x) is also such that x – 2 is always greater than or equal to 0 which implies that x is greater than or equal to 2. [Because we cannot put negative inside the square root]
The students must always mention the domain of the function always whenever talking about function because a function can never be judged without the domain.