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Homework Help: Even, odd or neither function

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose g is a function defined for all x > 0 which satisfies the following properties for all a, b > 0:

    g(1) = 0

    g(a/b) = g(a) - g(b)

    Determine whether the function f(x) = g(x+ root(x^2+1)) is even, odd or neither, and justify your answer.

    2. Relevant equations

    Odd function: f(-x) = -f(x)
    Even function: f(-x) = f(x)

    3. The attempt at a solution

    Well I was thinking that g could be a logarithmic function, and x+ root(x^2+1) is neither even nor odd, so f(x) is a neither function...

    I think I'm approaching this the wrong way... Can anyone help me out please?
  2. jcsd
  3. Sep 26, 2009 #2


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    Science Advisor

    The insight that g is a logarithmic function is good.

    Try solving f(x) + f(-x), and f(x) - f(-x). If f is odd or even, then one of these will have to reduce to zero.
  4. Sep 28, 2009 #3
    Ok so I took time to examine this problem again.

    g is a function defined for x>0, and satisfies the following properties for a,b > 0:

    g(1) = 0, g(a/b) = g(a) - g(b)

    Having seen that, I can assume that g is a logarithmic function because:

    let h be log x

    h is a function defined for x>0, and satisfies the following properties for a,b >0:

    h(1) = log 1 = 0, h(a/b) = log (a/b) = h(a) - h(b) = log a - log b

    So now I can assume g is a logarithmic function.

    Now I have to show if f(x) = g(x+ root((x^2)+1)) is even, odd, or neither.

    I'll check if f(x) is odd or not:

    f(-x) = f(x) <- definition of an odd function

    g(-x+root(((-x)^2)+1) = -g(x+root((x^2)+1))

    since g is a logarithmic function (from the assumption above, I can replace g with log(base 10))

    log(-x+root(((-x)^2)+1) = -log(x+root((x^2)+1))

    log(-x+root((x^2)+1) = -log(x+root((x^2)+1))

    the left hand side can not be explicitly interpreted whether it equals to -f(x) or not, so I have to rationalize the left hand side

    log[(-x+root((x^2)+1)(x+root((x^2)+1/x+root((x^2)+1)] = -log(x+root((x^2)+1))

    log(-x^2+x^2+1/x+root(x^2)+1)) = -log(x+root((x^2)+1))

    log(1/x+root(x^2+1) = -log(x+root((x^2)+1))

    from the algebraic properties of logarithms (log(x/y)=logx-logy), I can make the left hand side equal to:

    log 1 - log(x+root(x^2)+1)) = -log(x+root((x^2)+1))

    since g(1) = 0, log (1) = 0,

    0 - log(x+root(x^2)+1)) = -log(x+root((x^2)+1))

    - log(x+root(x^2)+1)) = -log(x+root((x^2)+1))

    now i replace log with g

    -g(x+root(x^2)+1)) = -g(x+root((x^2)+1))

    g(-x+root((-x)^2)+1)) = -g(x+root((x^2)+1)) (from above)

    therefore f(-x) = -f(x)

    so f(x) is an odd function


    Now can anyone confirm this please? ;) I apologize if this is hard to read, I do not know how to use latex coding...

    Much thanks in advance!
    Last edited: Sep 28, 2009
  5. Sep 29, 2009 #4
    Your solution (f(-x)=-f(x)) is correct. Just two remarks:
    1. You do not need to say that the function g is a logarithm (and it is not forcibly of base 10).
    2. You can proceed as you did to find the answer, that is, you start with f(-x)=-f(x) and try to reduce. But if you want to write it properly once you know it's true you have to start with f(-x) and you should arrive at -f(x).
    So my proposition would be:
    [tex] f(x) = g(x+\sqrt{x^2+1}). [/tex]
    We have
    [tex] f(-x) = g(-x+\sqrt{(-x)^2+1}) = g(-x+\sqrt{x^2+1}) = g\left(\frac{-x^2+x^2+1}{x+\sqrt{x^2+1}}\right) = g(1)-g(x+\sqrt{x^2+1}) = -f(x), [/tex]
    thus, [tex] f [/tex] is an odd function.
    If [tex] g [/tex] is zero everywhere [tex] f [/tex] is also an even function.
  6. Sep 29, 2009 #5
    Yes, moments after I posted this, I realized I did not have to replace g with a logarithm. Thanks for your input!
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