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Find Inverse of a Function Y=X2-X, X≥1/2

  1. Feb 13, 2015 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    Find the inverse of the function: Y=X2-X, X≥½

    2. Relevant equations


    3. The attempt at a solution

    I've switched X and Y to get: X=Y2-Y

    Then I've tried several things to get Y alone, but none of them seem to work. I've tried taking a square root, using a log, and several other things. I think using a log is the way to go here, but I can't seem to get it correct. I'm in the tutoring center here on campus and not even the tutors seem to be able to do it. I'd post my work, but I've gone through a half-dozen different ways of doing it and all of them come out to a dead end as far as I can tell, so I'm not sure what I'd post.

    Also, what's the significance of X≥½? Is it just extra info that I don't need? I know it limits using the function to only when X is equal to or greater than a half, but is that it?

    Thanks.
     
  2. jcsd
  3. Feb 13, 2015 #2

    pasmith

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    The general solution of [itex]x^2 + bx + c = 0[/itex] is [itex]x = \frac{-b \pm \sqrt{b^2 - 4c}}2[/itex].

    Which of [itex]\pm[/itex] should you take here, given that [itex]x \geq \frac12[/itex]?
     
  4. Feb 13, 2015 #3

    Dick

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    That's a quadratic equation in Y. Use the quadratic formula. Then you'll see what x>=1/2 has to do with the problem.
     
  5. Feb 13, 2015 #4

    LCKurtz

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    If you will complete the square on ##x^2-x## you will have ##y = (x-\frac 1 2)^2 - \frac 1 4##. This is a parabola opening up and is therefore not 1-1. But its vertex is at ##x=\frac 1 2## so the part with ##x>\frac 1 2## is increasing, therefore 1-1. Solve the completed square equation for ##x## taking the positive root to get ##f^{-1}(y)##.
     
  6. Feb 13, 2015 #5

    Drakkith

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    Sigh... silly quadratic forumula, always popping up where I least expect you...

    Okay, so I get Y=(1±√(1-4X))/2.

    I have no idea. How does that work with the inverse of a function? Does the X change to a Y here as well?
     
  7. Feb 13, 2015 #6

    Mark44

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    Now that you have switched the roles of x and y, any restrictions on x (such as x ≥ 1/2) now become restrictions on y. For this equation: $$y = \frac{1 \pm \sqrt{1 - 4x}}{2}$$
    which of the two branches produces y values that are at least 1/2?
     
  8. Feb 13, 2015 #7

    Mark44

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    @Drakkith, IMO the whole business of swapping x for y and y for x is highly overrated. Given an equation y = f(x), where f is a one-to-one function, you can find the inverse f-1 in one step: Solve for x to get x = f-1(y).

    It's only because we have a bias for making x the independent variable and y the dependent variable that we like to have functions in terms of x. That's where the x ↔ y business comes in. As you progress further toward calculus, switching variables becomes less important. For example, if one equation is y = x2 - 1, then you might have a reason to solve for x, which would be x = ± √(y + 1). If you have some information about x, you can decide which of the two square roots you should use.
     
  9. Feb 13, 2015 #8

    Drakkith

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    Since the right side is already 1 over 2, that would mean it would the radical would have to be positive, otherwise we'd subtract on the top and end up less than 1/2.
     
  10. Feb 13, 2015 #9

    Drakkith

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    Unfortunately I don't think my calculus teacher is going to agree. ;)
     
  11. Feb 13, 2015 #10

    Mark44

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    If you make a reasoned argument, I'm pretty sure he or she will agree (barring the possibility that your teacher is a moron ;), though college instructors often aren't).

    If you start with y = f(x) and do this:
    1. Solve for x

    you have x = f-1(y). (Again, assuming that f is one-to-one)

    If the question asks you for the inverse, as a function of x, then do the switcheroo.
     
  12. Feb 13, 2015 #11

    vela

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    You seem to have made a sign error somewhere. The square root isn't defined for X>1/4, but you know that the parabola opened upward so there should be a lower limit on X, not an upper limit.
     
  13. Feb 13, 2015 #12

    Drakkith

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    Ugh, the 1-4x should be 1+4x, shouldn't it?

    That makes it y=(1±√(1+4x))/2.
     
  14. Feb 13, 2015 #13

    Mark44

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    Hopefully you realize that this is not the inverse -- it's not even a function. You have to choose one of the branches based on the condition that y ##\ge## 1/2.
     
  15. Feb 13, 2015 #14

    Drakkith

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    Sure. It should be y=(1+√(1+4x))/2. Since the square root is always positive, having a 1- instead of 1+ would result in a value less than 1/2, which is outside the domain.
     
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