How do you find the inverse of #y=((x^2)-4)/x# and is it a function?

Graph the original function:

graph{(x^2-4)/x [-41.1, 41.1, -20.54, 20.55]}

Since it does not pass the horizontal line test, its inverse will not be a function.

Take the square root of both sides. Note that we will take the positive and negative versions of this -- this will actually create two separate functions that cannot act as a function on their own since together they break the vertical line test.

graph{(y-(x+sqrt(x^2+16))/2)(y-(x-sqrt(x^2+16))/2)=0 [-52.02, 52.02, -26, 26.02]}

To find the inverse of the function ( y = \frac{x^2 - 4}{x} ), we switch the roles of ( x ) and ( y ) and then solve for ( x ).

So, let ( x = \frac{y^2 - 4}{y} ).

Now, solve for ( y ) in terms of ( x ).

( xy = y^2 - 4 )

( y^2 - xy - 4 = 0 )

Using the quadratic formula, ( y = \frac{x \pm \sqrt{x^2 + 16}}{2} ).

So, the inverse function has two branches:

1. ( f^{-1}(x) = \frac{x + \sqrt{x^2 + 16}}{2} )
2. ( f^{-1}(x) = \frac{x - \sqrt{x^2 + 16}}{2} )

As for whether the inverse function is itself a function, it depends on the domain and range. Since we have two possible values of ( y ) for some ( x ) in the inverse function, it means the inverse function is not one-to-one, and therefore not a function.