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Math Help - Even And Odd Functions

  1. #1
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    Angry Even And Odd Functions

    Hey

    I just took a quiz and got this question wrong:

    Is the given functions even, odd, or neither?

    f(x) =  - 4x + \left| {8x} \right|

    I answered it was odd because the x coefficient of both terms have a power of 1:

    f(x) =  - 4{x^1} + \left| {8{x^1}} \right|

    Why did I get this answer wrong?

    Sam
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  2. #2
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    Re: Even And Odd Functions

    That is not the correct concept to used to determine if a function is even or odd. Do you know the definition for even and odd functions?
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    Re: Even And Odd Functions

    Quote Originally Posted by ArcherSam View Post
    Is the given functions even, odd, or neither?
    f(x) =  - 4x + \left| {8x} \right|
    f(2)=12~\&~f(-2)=20.
    If it were an even function then f(2)=f(-2)~.
    If it were an odd function then -f(2)=f(-2)~.
    Is it even or odd?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Even And Odd Functions

    General rules...

    a) f(x) is an even function if f(x)=f(-x)

    b) f(x) is an odd function if f(x)=-f(-x)

    c) any function f(x) can be written as f(x)=f_{e}(x)+f_{o}(x) where f_{e}(*) is the 'even part of f(*)' and f_{o}(*) is the 'odd part of f(*)'

    d) for any function f(x) is...

    f_{e}(x)= \frac{f(x)+f(-x)}{2}

    f_{o}(x)= \frac{f(x)-f(-x)}{2} (1)

    e) given f(x), one computes with (1) its even and odd part. If f_{o}(x)=0 then f(*) is an even function. If f_{e}(x)=0, then f(*) is an odd function. If f_{e}(x) \ne 0 and f_{o}(x) \ne 0, then f(*) in neither even nor odd...

    Kind regards

    \chi \sigma
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    Re: Even And Odd Functions

    Quote Originally Posted by cheme View Post
    That is not the correct concept to used to determine if a function is even or odd. Do you know the definition for even and odd functions?
    Sam, I would slightly edit this (^^) comment...

    If you have a polynomial with only even exponents (a constant "c" can be written as c*x^0, and zero is even, so constant terms are even...), then the function P(x) is even.
    If you have a polynomial with only odd exponents, then the function P(x) is odd.

    In your examples, there is an absolute value term, which is not polynomial.
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    Re: Even And Odd Functions

    Hello, Sam!

    I just took a quiz and got this question wrong:

    Is the given functions even, odd, or neither? .  f(x) \:=\: - 4x + |8x|

    I answered it was odd because the x of both terms have a power of 1:

    . . f(x) \:=\:-4{x^1} + |8x^1|

    Why did I get this answer wrong?

    Be careful!

    That rule about "all odd exponents" or "all even exponents"
    . . works with polynomials only. .
    (Edit: as TheChaz already pointed out.)

    If x^n is "inside" another function, all bets are off!
    Examples: . \sqrt{x}\quad e^x\quad|x|\quad\ln x


    Look at that function again: . f(x) \:=\:-4x + |8x|

    If x is negative (x = -a),
    . . we have: . f(-a) \:=\:-4(-a) + |8(-a)| \;=\;4a + 8a

    But -f(a) \:=\:-\left(-4a + |8a|\right) \:=\:-\(-4a + 8a) \:=\:4a - 8a

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    Re: Even And Odd Functions

    Quote Originally Posted by TheChaz View Post
    Sam, I would slightly edit this (^^) comment...

    If you have a polynomial with only even exponents (a constant "c" can be written as c*x^0, and zero is even, so constant terms are even...), then the function P(x) is even.
    If you have a polynomial with only odd exponents, then the function P(x) is odd.

    In your examples, there is an absolute value term, which is not polynomial.
    As my comment was directed towards the problem at hand, I should have said "the" function and not "a" function.
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  8. #8
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    Re: Even And Odd Functions

    @Soroban, cheme, TheChaz

    Thanks! I was wondering why it worked when my instructor used this method to determine if a function was even or odd. Now I know the using the powers of exponents to determine if a function is even or odd, only works when you have polynomials only.

    @Plato

    \begin{array}{l} f(x) = - 4x + \left| {8x} \right|\\ f( - x) = - 4( - x) + \left| {8( - x)} \right|\\ f( - x) = 4x + \left| { - 8x} \right|\\ f( - x) \ne f(x)\\ - f(x) = - (4x + \left| {8x} \right|)\\ - f(x) = - 4x - \left| {8x} \right|\\ - f(x) \ne f(x)\\ neither \end{array}

    @chisigma

    I was trying to understand general rule c, and d. May you solve a problem using rule c and d.
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  9. #9
    MHF Contributor chisigma's Avatar
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    Re: Even And Odd Functions

    Quote Originally Posted by ArcherSam View Post
    @Soroban, cheme, TheChaz

    Thanks! I was wondering why it worked when my instructor used this method to determine if a function was even or odd. Now I know the using the powers of exponents to determine if a function is even or odd, only works when you have polynomials only...

    @chisigma

    I was trying to understand general rule c, and d. May you solve a problem using rule c and d.
    All right!... is...

    f(x)= f_{e}(x) + f_{o}(x) = -4 x + |8 x| (1)

    ... so that...

    f_{e}(x)= \frac{f(x)+f(-x)}{2}= {-4 x + |8 x| +4 x + |8 x|}{2} = |8 x| (2)

    f_{o}(x)= \frac{f(x)-f(-x)}{2}= {-4 x + |8 x| -4 x - |8 x|}{2} = -4 x (3)

    Because neither f_{e}(*) nor f_{o}(*) is 'identically 0' , then f(x) is 'neither even nor odd'...

    If You will use this approach in the future, You will be 'very lucky' because it works for any type of fucntion, not only for polynomials...

    Kind regards

    \chi \sigma
    Last edited by chisigma; October 13th 2011 at 12:34 AM.
    Thanks from ArcherSam
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