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I Pre Image

  1. Jan 12, 2017 #1
    I'm trying to understand image and pre image better but I am having a hard time finding good examples.

    So here is one I did come across, let's say ##f:\mathbb{R} →\mathbb{R}## defined by ##f(x)=x^{2}##. Suppose also that ##D = [-1,2]## where ##D\subset \mathbb{R}##. If I'm looking for ##f^{-1}(D)## then I can only use ##D\setminus [-1,0)## in which case ##f^{-1}(D)=[-\sqrt{2},\sqrt{2}]##correct?
     
    Last edited: Jan 12, 2017
  2. jcsd
  3. Jan 12, 2017 #2
    Not correct. Rather, ##f^{-1}(D) = [-\sqrt{2}, \sqrt{2}]##, since ##f^{-1}(D)## is defined as the set of all ##x## such that ##f(x) \in D##. Also you can use ##D## and the inverse image is always well defined, even if ##D## is not completely contained in the range ##f(\mathbb{R})##. In fact, ##f^{-1}((-\infty, 2]) = [-\sqrt{2}, \sqrt{2}]##.

    P.S. You really want double hashtags to begin and end a latex formula, rather than code
     
  4. Jan 12, 2017 #3
    Can you explain why ## f^{-1}(D) = [0,\sqrt{2}]## ? Why would we not include ##[-\sqrt{2},0]## in our pre-image? Explain please.
     
  5. Jan 12, 2017 #4
    My bad, I've made a mistake. Corrected it.
     
  6. Jan 12, 2017 #5
    Okay, so what if we wanted to find ##f^{-1}(F)## using the same function with ##F=[-4,-1]##, in that case we would have ##∅##.
     
  7. Jan 12, 2017 #6
    Yes. Unless of course you decide to change domains to complex numbers.
     
  8. Jan 12, 2017 #7
    Of course :)
     
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