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Inverse & preimage

  1. Oct 23, 2008 #1
    1) Consider the example y=f(x)=x3
    My statistics textbook say x=f -1(y)=y1/3 is the inverse of f
    On the other hand, my calculus textbook says y=x1/3 is the inverse of f
    So I am confused...it looks like the idea of inverse is used inconsistently. (When you plot both functions on the xy-plane, you will certainly see two different graphs.)
    Which one is the correct one according to the precise definition of inverse?


    2) I don't get the difference between the "inverse" of f and the "inverse image" or "premiage". Can somebody explain?


    Thank you!
     
  2. jcsd
  3. Oct 24, 2008 #2
    Note in this example that f-1 is written as a function of y instead of x. That's where your confusion is coming from. The inverse function could (and for clarity's sake, probably should) be written as [tex]f^{-1}(x)=x^{1/3}[/tex].


    Suppose f is a function. Another function f-1 is called the inverse of f if [tex]f(f^-1(x)) = f^-1(f(x)) = x[/tex] for all x. In other words, an inverse function is just a function with a special relation to another.

    The preimage of a point under a function is a the set of points which map to that point. In other words preimage(p) = {x such that f(x) = p}. So the preimage of a point is a set.
     
  4. Oct 24, 2008 #3

    HallsofIvy

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    By the way, even if a function, f, does not have an inverse, we can still define the "inverse image", f-1(A). For example, if f(x)= x2, there is no "f-1(x)" because f is not "one-to-one"; since f(2)= 4 and f(-2)= 4, which would be f-1(4)?

    But if B= [0, 1], we can still have f-1(B)= [-1, 1] since, for any x in [-1, 1], f(x)= x2 is in [0,1].

    I once made a fool of myself, presenting a proof in a graduate class, by forgetting that! I was to prove a statement about inverse images and did it assuming the function f must have an inverse function.
     
  5. Oct 24, 2008 #4
    Ah yes! This is an important point!

    We can also "map" functions over sets. So if A is a set of numbers and f is a function, then we define f(A) = {f(x) for each x in A}.

    This gives us cute little properties like

    [tex]x \in f(A) \Leftrightarrow f^{-1}(x) \in A[/tex]
     
  6. Oct 25, 2008 #5
    Thanks! Now I have an idea of the difference between inverse & premiage.

    Back to 1) y=f(x)=x^3
    Is it even correct to say that x=f -1(y)=y^(1/3) is the inverse of f ? My statsitics textbook is keep doing the same thing again and again...but then there would be inconsistency...x=y^(1/3) and y=x^(1/3) do not give the same graph when you graph them on the xy-plane.
     
  7. Oct 25, 2008 #6

    HallsofIvy

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    No they don't. But one is "x as a function of y" and the other is "y as a function of x". It is still the SAME function in both formulas.
     
  8. Oct 26, 2008 #7
    Standard math notation makes this slightly more confusing than it needs to be.

    When you have a function defined as [tex]f(x) = x^3[/tex], the function itself is named "f". You would say "f is a cubic function". However, it is very common to confuse "f" with "f(x)" and say that "f(x) is a cubic function". Technically f(x) means "f evaluated at x" or "f with the argument x supplied to it" or something, but in practice, it's usually clear from context what you mean.

    A related consequence is that the variable name DOES NOT MATTER. It is arbitrary. If [tex]f(x) = x^3[/tex], then it is just as true to say [tex]f(y) = y^3[/tex]. Just like how [tex]\Sigma_{i=0}^\infty \frac{1}{2^i}[/tex] is the same as [tex]\Sigma_{k=0}^\infty \frac{1}{2^k}[/tex]. j and k are just dummy variables. Variables used in the definition of functions are the same.

    The confusion comes from the implicit function theorem. That is, whenever you have an equation like "y = 3+x" or "z = x + y + z", you can define a function in a natural way. In the first case, "y = 3+x", you could define a function f(x) = 3 + x. In the second, you could define a function of three variables. It is a technique used so often that people almost always start confusing y, which is a real number, and f, which is a function.
     
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