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Odd or even function?

  1. Aug 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine whether the following function is odd or even or none:
    $$f(x)=\log \left(x+\sqrt{1+x^2} \right)$$

    2. Relevant equations

    3. The attempt at a solution
    For an even function, ##f(-x)=f(x)## and for an odd function ##f(-x)=-f(x)##.
    Replacing x with -x in the given function,
    I don't see it being equal to f(x) or -f(x) so it should be neither odd nor even but the answer key states it is an odd function. :confused:

    Any help is appreciated. Thanks!
  2. jcsd
  3. Aug 6, 2013 #2
    hint: simplify the argument.
  4. Aug 6, 2013 #3
    The only thing I can think of is multiplying and dividing by ##\sqrt{1+x^2}+x##, do you ask me this?
  5. Aug 6, 2013 #4


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    Why don't you just try it and see?
  6. Aug 6, 2013 #5
    Woops, just tried that and it does come out to be -f(x). Thanks dirk_mec1! :tongue2:
  7. Aug 6, 2013 #6


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    I like this question, very nice it is.

    Hint: compare f(x) and f(-x), what do you see inside the logs? They are begging to have something done to them...

    You've solved it now I see, so I'll just mention it. They are conjugate expressions that deserve to be multiplied together.
  8. Aug 6, 2013 #7
    Multiplied? Or should I add them?
  9. Aug 6, 2013 #8


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    Multiply the expressions inside the logs, which is the same as adding the logs.
    Going back to the problem as given, the most natural thing to try is f(x)+f(-x). As soon as you see the form log(..)+log(..), the next step should be automatic.
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