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Help with the inverse of some functions

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data

    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.

    3. 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
     
  2. jcsd
  3. Jan 20, 2015 #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?
     
  4. Jan 20, 2015 #3
    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).
     

    Attached Files:

  5. Jan 20, 2015 #4

    Orodruin

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  6. Jan 20, 2015 #5
    Honestly, I only used before the quadratic formula, but with its derivation I was not familiarized.
     
  7. Jan 20, 2015 #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.
     
  8. Jan 20, 2015 #7

    Ray Vickson

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    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.
     
  9. Jan 20, 2015 #8
    So you are telling me that I can solve it by using the quadratic formula and finding the valuex of x?
     
  10. Jan 20, 2015 #9

    Orodruin

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    Yes.
     
  11. Jan 20, 2015 #10

    statdad

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    What are the domains? if the function is not 1-1 there will not be a unique inverse.
     
  12. Jan 20, 2015 #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
     
  13. Jan 20, 2015 #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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