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The *reciprocal* of a one-to-one function

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    If g(x) is one-to-one and differentiable at every point in R, then its reciprocal 1/g(x) is also differentiable at every point in its domain. True of False?

    2. Relevant equations



    3. The attempt at a solution

    I have tried several times, and the statement seems to be right. I think 1/g(x) can be regarded as 1/u and u=g(x), so 1/g(x) ' = 1/u ' * g'(x)=g'(x)/g(x) (I don't know if this is right...), but I'm stucked here and can't go any further because I don't know how to prove a function to be one-to-one..

    I'll be very grateful if anyone can help me.. Thanks!!!
     
    Last edited: Oct 12, 2011
  2. jcsd
  3. Oct 12, 2011 #2

    Dick

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    (1/g(x))'=g'(x)/g(x) is wrong. Can you correct it? The only thing I can think of that would involve 'one-to-one' is if you have stated the problem wrong. Did it ask if the INVERSE FUNCTION g^(-1)(x) is differentiable, not the reciprocal (g(x))^(-1) or 1/g(x)?
     
  4. Oct 12, 2011 #3
    Hi, thank you so much. I calculated once again: g^2(x)'/g^2(x) Is this one right?

    It is the reciprocal, the 1/g(x). The question asks for if 1/g(x) is differentiable at every point in its domain.
     
  5. Oct 12, 2011 #4

    Dick

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    No, you haven't got it right yet. Use the quotient rule.
     
  6. Oct 12, 2011 #5
    Is it -g'(x)/g^2(x) ?
     
  7. Oct 12, 2011 #6

    Dick

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    Yes it is! Now you know g'(x) exists for all x. So the quotient -g'(x)/g(x)^2 exists everywhere that g(x) is not equal to zero, right? Are points where g(x)=0 in the domain of 1/g(x)? I'm still not seeing what this has to do with g(x) being one-to-one.
     
  8. Oct 12, 2011 #7

    Bacle2

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    But what happens at points where f is 0?
     
  9. Oct 12, 2011 #8

    Dick

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    What's f??
     
  10. Oct 12, 2011 #9

    Bacle2

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    Oh, I meant g.
     
    Last edited: Oct 12, 2011
  11. Oct 12, 2011 #10
    Does one-to-one means that g(x) can't be 0? Since if g=0 then it can't pass the horizontal line test.
     
  12. Oct 12, 2011 #11

    Bacle2

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    g(x)=x is 1-1.
     
  13. Oct 12, 2011 #12

    Dick

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    Not at all. g(x)=x is one-to-one. And g(0)=0. As I said, I don't get what one-to-one has to do with this problem if it's reciprocal.
     
  14. Oct 12, 2011 #13
    Oh I got it. Thank you so much!
     
  15. Oct 13, 2011 #14

    HallsofIvy

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    The constant function g(x)= 0 for all x is not one- to- one. Is that what you were thinking of?
     
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