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Math Help - Even and Odd Functions

  1. #1
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    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
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    Be careful with your notation. Exponentiation takes precedence over negation, so -x^2 is usually interpreted -(x^2), not (-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 -f(x)=f(-x) (which is what you are trying to show). To show that f is odd, you should do

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

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

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

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

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

    =f(-x).
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