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)

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

 

[Orodruin]

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

 

[saulwizard1]

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).

 

[Orodruin]

Are you familiar with the quadratic formula and its derivation?
http://en.wikipedia.org/wiki/Quadratic_formula

 

[saulwizard1]

Are you familiar with the quadratic formula and its derivation?
http://en.wikipedia.org/wiki/Quadratic_formula

Honestly, I only used before the quadratic formula, but with its derivation I was not familiarized.

 

[RUber]

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.

 

[Ray Vickson]

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)

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.

 

[saulwizard1]

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?

 

[Orodruin]

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

Yes.

 

[statdad]

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

 

[saulwizard1]

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

 

[saulwizard1]

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))