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Graph problem

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]


    2. Relevant equations
    Not necessary


    3. The attempt at a solution
    The answers that I have up there are correct, but I was wondering why graph 2 did not look something like this:
    [​IMG]
     
  2. jcsd
  3. Feb 5, 2008 #2

    rock.freak667

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    Homework Helper

    Let y=x^2 so x=sqrt(y)

    so [itex]f^{-1}(x)=\sqrt{x}[/itex] and [itex]f^{-1}(f(x))=\sqrt{x^2}[/itex]
     
  4. Feb 6, 2008 #3
    In fact, the inverse of [itex]y=x^2[/itex] is [itex]y=\sqrt{x}, x \geq 0[/itex] and [itex]y=-\sqrt{x}, x < 0[/itex]. It doesn't matter in the end, however, because the definition of an inverse says that [itex]f^-1(f(x)) = x[/itex]
     
  5. Feb 6, 2008 #4

    HallsofIvy

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    Actually, that's a seriously bad question! Since y= x2 is not one-to-one, it does not have an inverse! temaire, the graph you showed would be the "inverse function" to x2 but, as you can see from the graph, there are two values of x for each positive y and so it is not a function.,
     
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