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Finding inverse when function conatains absolute value

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Given the function f(x) = (abs(x))*x +6, find f^-1(x)

    2. Relevant equations



    3. The attempt at a solution

    for x≥ 0, f(x) = x^2 + 6
    y=x^2 +6
    x = √(y-6) for y≥6
    → f^-1(x) = √(x-6) for x≥6

    for x< 0, f(x) = -x^2 + 6
    y= -x^2 +6
    x = √(6-y) for y<6
    → f^-1(x) = √(6-x) for x<6

    But the correct answer is
    f^-1(x) = ±√(x-6) with the domain and range as all real numbers.
    How do I get the inverse function to not be piecewise?
    Sorry about the lack of proper equations, I don't know how to use latex.
     
  2. jcsd
  3. Nov 17, 2013 #2

    462chevelle

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    have to restrict its not a 1 to 1
     
  4. Nov 17, 2013 #3
    Could you please elaborate? How is the function not one to one? f(x) = (abs(x))*x +6 results in a graph that essentially looks like a cubic function due to the absolute value sign.
     
  5. Nov 17, 2013 #4

    462chevelle

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    alright. sorry, the way it was written caused me to jump the gun.
    if you just solve it without the restrictions on x that's what you get.
    but other than that im not sure. will have to wait on someone smarter.
     
  6. Nov 17, 2013 #5

    LCKurtz

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    You should have ##x =\pm\sqrt{6-y}## and you need to choose the - sign. So you would have
    ##f^{-1}(x) = -\sqrt{6-x},~x<6##.

    Their "correct answer" is wrong. Your two piece answer is correct once you fix that minus sign.
     
  7. Nov 17, 2013 #6
    Yes, of course that makess sense regarding the minus sign. Oops. The reason why I specified that particular answer, ie ± √(x-6) is because that is what WolframAlpha gives as the result to the inverse of the original function. I can't think of how to arrive to that result.
     
  8. Nov 17, 2013 #7

    462chevelle

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  9. Nov 17, 2013 #8
    It's confusing me, because that's the book answer and the wolfram answer! There must be some arithmetic trick.
     
  10. Nov 17, 2013 #9

    LCKurtz

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    If you are confused, draw a graph of your original ##f(x)## and your ##f^{-1}(x)## on the same picture. If your inverse is the reflection of the original in the ##45^\circ## line, yours is correct. That's all there is to it.
     
  11. Nov 17, 2013 #10

    462chevelle

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  12. Nov 17, 2013 #11

    LCKurtz

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    What's interesting about that? Neither is a complete and correct picture of what I suggested in post #9. Can't you just do it by hand picking a few nice numbers?
     
  13. Nov 17, 2013 #12

    462chevelle

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    I guess if youre just plugging numbers in for y and solving for x with both -(6-x)^1/2 and (x-6)^1/2
    you get the same x value.
    nevermind that. I keep leaving out stupid details.
    why are wolfram and the book both wrong? is there something else to consider?
     
    Last edited: Nov 17, 2013
  14. Nov 17, 2013 #13

    LCKurtz

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    They are wrong because they are wrong, what else is there to say. In post #2 you described the graph of ##f(x)##. In post #5 I showed you a sign correction so you have ##f^{-1}(x)##. As I said on post #9, graph them both on the same graph. Do it by hand if you can't make Wolfram behave with two piece functions.
     
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