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Homework Help: F(x) function help

  1. May 11, 2010 #1
    Given [tex]f(x)=x^2-4x[/tex] , [tex]x\in R[/tex] , [tex]|x|\leq 1[/tex]

    so f(x) is a one to one function , and i am supposed to find its inverse .

    and i found :

    [tex]f^{-1}(x)=2\pm \sqrt{4+x}[/tex]

    this is weird since a single input would give 2 different outputs and it cant be considered a function . But f(x) is a one-one function , so it should have an inverse ?
     
  2. jcsd
  3. May 11, 2010 #2

    Cyosis

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    Homework Helper

    Re: function

    You have solved the equation [itex]y=x^2-4x[/tex] for all x in R. Yet the original function only exist for |x|<=1. This restricts the domain such that it is one to one (it is not one to one for all x in R). Use this domain to determine which branch you need.
     
  4. May 11, 2010 #3
    Re: function

    thanks ,i thought so too but i am not sure how to see from the domain that the inverse should be 2+root(4+x) OR 2-root(4+x)
     
  5. May 11, 2010 #4

    Mark44

    Staff: Mentor

    Re: function

    The domain of f is {x | |x| <= 1}. That domain can be written as an interval, which should help you figure out which branch to use for the inverse.
     
  6. May 11, 2010 #5

    Cyosis

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    Re: function

    The range of f is the domain of f^-1. If f(a)=b then f^-1(b)=a.
     
  7. May 11, 2010 #6
    Re: function

    thanks again !

    so the domain of f(x) is between 1 and -1 , and the range is between -3 and 5 , so the domain of f^(-1)(x) is also between -3 and 5 ? so both +ve and -ve would produce such range , err i am still confused ,
     
  8. May 11, 2010 #7

    Cyosis

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    Re: function

    I suggest you plug in some numbers to see what happens.
     
  9. May 11, 2010 #8
    Re: function

    thanks !!
     
  10. May 11, 2010 #9

    D H

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    Staff Emeritus
    Science Advisor

    Re: function

    Broadening things out, it depends on what you mean by one-to-one. The term one-to-one means injective to some, bijective to others. The function is a one-to-one function by both meanings of the term over the domain |x|≤1. It is not bijective over all of the reals, so it does not have an inverse function for this extended domain. (It does however have an inverse multivalued function.)
     
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