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Mistake on other thread, Here is the problem:

  1. Oct 10, 2006 #1
    Ok sorry about that, here is the actual problem:

    g(x)=sq.rt.(2-x)

    Find the inverse of g(x)

    ok the answer is -x^2+2, but how come the domain is [0,infinity)??
     
  2. jcsd
  3. Oct 10, 2006 #2
    The range of [tex] g(x) [/tex] is equaled to the domain of [tex] g^{-1}(x) [/tex]. The range of [tex] g(x) [/tex] is [tex] [0, \infty) [/tex]. Thus the domain of [tex] g^{-1}(x) [/tex] is [tex] [0, \infty) [/tex]. Or think of it like this: all the x-values of [tex] g^{-1}(x) [/tex] are really all the y-values of [tex] g(x) [/tex].
     
  4. Oct 10, 2006 #3
    Oh it all makes sense now, thanks! :smile:
     
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