# Help with the inverse of some functions

• saulwizard1

## Homework Statement

Hi!
Does anyone know how to solve the inverse of these functions?
• y=(4x^2+2x-2)/(8x^2-4x+6)
• y=(x+1)/(x^2)

Sincerely, I would appreciate your help with these exercises.

## The Attempt at a Solution

For the first one: 8yx^2-4xy+6y=4x^2+2x-2

For the second exercise:
yx^2=x+1
yx^2-x=1

So you now have a second order polynomial in x. How do you solve this if you assume y to be fixed?

So you now have a second order polynomial in x. How do you solve this if you assume y to be fixed?
I've tried to solve it by using a factorization, but I get lost on the next steps, and I don't know how to do it exactly. I attach a file with my attempts (I don´t know if I am correct or not).

#### Attachments

• inverse functions.docx
12.9 KB · Views: 574
Both of these reduce to quadratic equations.
When I invert functions, I normally like to swap my x and y variables and solve for y in terms of x. It is just the same to solve for x in terms of y, but old habits die hard, and I am used to looking for functions of x.
If you do this, you will be able to solve for y by treating functions of x as coefficients (A =f(x) , B=g(x), C=h(x) ) and using the quadratic equation.

## Homework Statement

Hi!
Does anyone know how to solve the inverse of these functions?
• y=(4x^2+2x-2)/(8x^2-4x+6)
• y=(x+1)/(x^2)

Sincerely, I would appreciate your help with these exercises.

## The Attempt at a Solution

For the first one: 8yx^2-4xy+6y=4x^2+2x-2

For the second exercise:
yx^2=x+1
yx^2-x=1

What is stopping you from solving the quadratic equations, to find x in terms of y?

Of course, each quadratic will have two roots (at least for most values of y) so you will have two values of x for each y, and you need to worry about what that means.

What is stopping you from solving the quadratic equations, to find x in terms of y?

Of course, each quadratic will have two roots (at least for most values of y) so you will have two values of x for each y, and you need to worry about what that means.
So you are telling me that I can solve it by using the quadratic formula and finding the valuex of x?

So you are telling me that I can solve it by using the quadratic formula and finding the valuex of x?
Yes.

What are the domains? if the function is not 1-1 there will not be a unique inverse.

Ok, I think I have the answer for the second exercise, and now I share with you
y(x)=(1+-sqrt(4x+1))/2x

Now I share the answer for the first exercise, I expect don't have mistakes.

y(x) = (2 x+1)/(4 (2 x-1))±sqrt(-44 x^2+12 x+9)/(4 (2 x-1))