How do you determine whether the function \[f\left( x \right) = \frac{1}{{{x^2}}}\] has an inverse and if it does, how do you find the inverse function?

Hint: For determining the function \[f\left( x \right) = \frac{1}{{{x^2}}}\] has an inverse or not. In order to find the inverse of the given function \[f\left( x \right) = \frac{1}{{{x^2}}}\], interchange the variable ‘x’ with the variable ‘y’ and then solve for the variable ‘y’.

Complete step-by-step solution:
The given function we have \[f\left( x \right) = \frac{1}{{{x^2}}}\] ,
In order to find the inverse of the given function \[f\left( x \right) = \frac{1}{{{x^2}}}\], interchange the variable ‘x’ with the variable ‘y’ and then solve for the variable ‘y’.
Therefore, we get the following result,
$
   \Rightarrow f\left( x \right) = \frac{1}{{{x^2}}} \\
   \Rightarrow y = \frac{1}{{{x^2}}} \\
$
Now, exchange the variable, we will get ,
$ \Rightarrow x = \frac{1}{{{y^2}}}$
Now, solve for variable ‘y’ , we will get ,
$
   \Rightarrow {y^2} = \frac{1}{x} \\
   \Rightarrow y = \pm \frac{1}{{\sqrt x }} \\
$
The inverse of the given function is not a function because it does not pass the vertical line test. However, there exist two methods to restrict its domain so that the existing inverse is a function.

Method I : We have to restrict the domain of the given function as \[x \geqslant 0\]. Then, the range of the inverse of the function is given as \[y \geqslant 0\]. Since the inverse of the function passes the vertical line test, it represents a function.

Method II : We have to restrict the domain of the given function as \[x \leqslant 0\]. Then, the range of the inverse of the function is given as \[y \leqslant 0\]. Since the inverse of the function passes the vertical line test, it represents a function.

Note: If we have questions similar in nature as that of above can be approached in a similar manner and we can solve it easily and can find the corresponding result. We just interchange the variable ‘x’ with the variable ‘y’ and then solve for the variable ‘y’.