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Thread: Even and Odd Functions

  1. #1
    Feb 2009

    Even and Odd Functions

    I am correcting my tests for a little review for my final on Thursday and got hung up on this question I got wrong when I took it.

    Here goes nothing:

    Algebraically determine whether the following function is odd, even or neither.
    F(x) = (sin(x)/(x^2)+5) - tan(x)

    f(-x) = (sin(-x)/(-x^2)+5) - tan(-x)
    f(-x) = (-sin(x)/(x^2)+5) + tan(x)

    f(-x) does not equal f(x) so it cant be equal, now to check if its odd.
    if -(f(x)) = f(-x), it will be odd.

    -(f(x)) = -((-sin(x)/(x^2)+5) +tan(x))
    -(f(x)) = (sin(x)/(-x^2)-5) - tan(x))
    -(f(x)) = (sin(x)/(x^2)-5) - tan(x)

    Thus -(f(x)) does NOT equal f(-x).
    So its neither.

    Or am I missing something.
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  2. #2
    MHF Contributor Reckoner's Avatar
    May 2008
    Baltimore, MD (USA)


    Be careful with your notation. Exponentiation takes precedence over negation, so $\displaystyle -x^2$ is usually interpreted $\displaystyle -(x^2),$ not $\displaystyle (-x)^2.$

    Quote Originally Posted by Byud View Post
    if -(f(x)) = f(-x), it will be odd.

    -(f(x)) = -((-sin(x)/(x^2)+5) +tan(x))
    -(f(x)) = (sin(x)/(-x^2)-5) - tan(x))
    -(f(x)) = (sin(x)/(x^2)-5) - tan(x)
    I am not sure what you are doing here, as you start off by assuming that $\displaystyle -f(x)=f(-x)$ (which is what you are trying to show). To show that $\displaystyle f$ is odd, you should do

    $\displaystyle f(x)=\frac{\sin x}{x^2+5}-\tan x$

    $\displaystyle \Rightarrow-f(x)=-\left(\frac{\sin x}{x^2+5}-\tan x\right)$

    $\displaystyle =-\frac{\sin x}{x^2+5}+\tan x$

    $\displaystyle =\frac{-\sin x}{x^2+5}+\tan x$

    $\displaystyle =\frac{\sin(-x)}{(-x)^2+5}-\tan(-x)$

    $\displaystyle =f(-x).$
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