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Help with the inverse of some functions

  • #1

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
 

Answers and Replies

  • #2
Orodruin
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So you now have a second order polynomial in x. How do you solve this if you assume y to be fixed?
 
  • #3
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

  • #5
  • #6
RUber
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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.
 
  • #7
Ray Vickson
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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.
 
  • #8
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?
 
  • #9
Orodruin
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So you are telling me that I can solve it by using the quadratic formula and finding the valuex of x?
Yes.
 
  • #10
statdad
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What are the domains? if the function is not 1-1 there will not be a unique inverse.
 
  • #11
Ok, I think I have the answer for the second exercise, and now I share with you
y(x)=(1+-sqrt(4x+1))/2x
 
  • #12
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))
 

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