Inverse Functions Help?

Question

Find the inverse of the following function where the domain of the function is x > 0. Leave your answer in terms of y.

Y= x² –1 / x

Answer 1

y = (x^2 - 1)/x

To find the inverse, swap the x and y variables, and then solve for y.

x = (y^2 - 1)/y

Multiply both sides by y

xy = y^2 - 1

Bring everything to the right hand side.

0 = y^2 - xy - 1

Now, we want to solve this using the quadratic formula. In this case, a = 1, b = -x (the coefficient of y), and c = -1. Therefore

y = [1 +/- sqrt[ (-x)^2 - 4(-1) ] ] / 2
y = [1 +/- sqrt(x^2 + 4)] / 2

Note that, at this point, this is no longer a function. However, we're given the domain of the original function was that x > 0.
One noteable trait about functions and inverses is that they swap domains and ranges, so it follows that the inverse of this function will have a range of y > 0.

To summarize, we have two functions:

y = [1 + sqrt(x^2 + 4)] / 2

There's no doubt that y > 0 in this case.

y = [1 - sqrt(x^2 + 4)] / 2

This case is questionable, however. This cannot happen, and we can prove this by contradiction.
Assume that this can happen. Then

y > 0, so
[1 - sqrt(x^2 + 4)] / 2 > 0.
Multiply both sides by 2,

1 - sqrt(x^2 + 4) > 0

Moving the 1 to the other side,

-sqrt(x^2 + 4) > -1

Now, multiplying both sides by (-1) and flipping the inequality,

sqrt(x^2 + 4) < 1.

Square both sides,

x^2 + 4 < 1, and
x^2 < -3, which is impossible because x^2 is always greater than or equal to 0.

That means we can reject the possibility that

y = [1 - sqrt(x^2 + 4)] / 2

And as a result, our inverse is

y = [1 + sqrt(x^2 + 4)] / 2

Answer 2

x = y^2 - 1/y

Answer 3

y=x^3 i think