Question

Which of the following functions is an inverse function?
A) $f\left( x \right) = \dfrac{1}{{x - 1}}$
B) $f\left( x \right) = {x^2}$ for all $x$
C) $f\left( x \right) = {x^2},x \geqslant 0$
D) $f\left( x \right) = {x^2},x \leqslant 0$

Answer 1

A function is said to be an inverse function if it undoes the action of some other function. For a function to be an inverse function, it should be a one-to-one function. So check whether the given functions are one-to-one or not.

Complete step by step answer:
Let us check for these functions individually.
A) Given function is $f\left( x \right) = \dfrac{1}{{x - 1}}$
In order to check whether this function is one-to-one or not, we need to prove that if $f\left( x \right) = f\left( y \right)$ then $x = y$
So let us start with $f\left( x \right) = f\left( y \right)$
By substituting these values in the given function, we get $\dfrac{1}{{x - 1}} = \dfrac{1}{{y - 1}}$
By solving, we get $x - 1 = y - 1 \Rightarrow x = y$
This means that the function $f\left( x \right) = \dfrac{1}{{x - 1}}$ is a one-to-one function proving that it is an inverse function.

For option B, $x$ will have two values when the square is removed so it cannot be one-to-one.

C) Given function is $f\left( x \right) = {x^2},x \geqslant 0$
Similarly, we write $f\left( x \right) = f\left( y \right)$ for this function.
By substituting, we get ${x^2} = {y^2}$
Generally, if we remove the square of this equation $x$ would have two values of $y$ i.e. $\pm y$ but in this case it is given that the value of $x$ is greater than or equal to zero.
So the equation will be $x = y$
Therefore this function is one-to-one and hence is an inverse function.

For option D, it is given that $x$ is less than or equal to zero. This means that the inverse function would have only imaginary values so this function cannot be an inverse.
Therefore option A and C are inverse functions.

Note: A one-to-one function can be explained as follows:
If a function $f\left( x \right) = y$ then for every value of $y$ there should be a unique value of $x$ in the co-domain of the function.