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Thread: Functions: Even, Odd or Neither

  1. #1
    Super Member harpazo's Avatar
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    Functions: Even, Odd or Neither

    Determine algebraically if each function is even, odd, or neither.

    1. F(x) = cuberoot{x}

    2. F(x) = 2x/|x|

    3. f(x) = x + |x|

    Like always, seeking the needed steps.
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    MHF Contributor MarkFL's Avatar
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    Re: Functions: Even, Odd or Neither

    A function is even if:

    $\displaystyle f(-x)=f(x)$

    A function is odd if:

    $\displaystyle f(-x)=-f(x)$

    Using these rules, what do you conclude about the given functions?
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    Re: Functions: Even, Odd or Neither

    An example:
    \begin{align}\text{let} \quad f(x) &= x^2 \\ \text{then} \quad f(-x) &= (-x)^2 \\ &= \big((-1)x\big)^2 \\ &= (-1)^2 x^2 \\ &= (1)x^2 \\ &= x^2 \\ &= f(x) \end{align}
    So $f(x)$ Is even.

    You can follow similar procedures for your functions.
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    Re: Functions: Even, Odd or Neither

    can we at least assume that you know that |x|= x if $\displaystyle x\ge 0$ and |x|= -x if x< 0.

    From that
    2) $\displaystyle f(x)= \frac{2x}{|x|}$
    if $\displaystyle x\ge 0$ $\displaystyle f(x)= \frac{2x}{x}= 2$ and if $\displaystyle x< 0$, $\displaystyle f(x)= \frac{2x}{-x}= -2$.

    3) $\displaystyle f(x)= x+ |x|$
    if $\displaystyle x\ge 0$ $\displaystyle f(x)= x+ x= 2x$ and if $\displaystyle x< 0$, $\displaystyle f(x)= x+ (-x)= 0$.
    Last edited by topsquark; Dec 14th 2018 at 09:36 AM. Reason: Tweaked LaTeX
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    Re: Functions: Even, Odd or Neither

    Quote Originally Posted by harpazo View Post
    Determine algebraically if each function is even, odd, or neither.
    2. $F(x) = \dfrac{2x}{|x|}$
    Like always, seeking the needed steps.
    $\displaystyle \begin{align*}F(-x) &= \dfrac{-2x}{|-x|} \\&= -\dfrac{2x}{|x|} \\&= -F(x)\end{align*}$
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    Super Member harpazo's Avatar
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    Re: Functions: Even, Odd or Neither

    Quote Originally Posted by MarkFL View Post
    A function is even if:

    $\displaystyle f(-x)=f(x)$

    A function is odd if:

    $\displaystyle f(-x)=-f(x)$

    Using these rules, what do you conclude about the given functions?
    1. F(x) = cuberoot{x}

    F(-x) = cuberoot{-x}

    Result: neither

    2. F(x) = 2x/|x|

    F(-x) = 2(-x)/|-x| = -2x/x = -2

    Result: neither

    3. f(x) = x + |x|

    f(-x) = -x + |-x| = -x + x = 0

    Result: neither
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    Forum Admin topsquark's Avatar
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    Re: Functions: Even, Odd or Neither

    Quote Originally Posted by harpazo View Post
    1. F(x) = cuberoot{x}

    F(-x) = cuberoot{-x}

    Result: neither

    2. F(x) = 2x/|x|

    F(-x) = 2(-x)/|-x| = -2x/x = -2

    Result: neither

    3. f(x) = x + |x|

    f(-x) = -x + |-x| = -x + x = 0

    Result: neither
    Careful! |-x| is not equal, in general, to x.

    $\displaystyle |x| = \begin{cases} -x & x < 0 \\ x & 0 < x \end{cases}$

    We don't know if x is negative or positive. So the only thing you can say is
    $\displaystyle F(-x) = \dfrac{2(-x)}{|-x|} = \dfrac{-2x}{|x|} = -\dfrac{2x}{|x|} = -F(x)$

    As HallsofIvy and Plato have already told you.

    You have a similar problem with #3.


    -Dan
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    MHF Contributor MarkFL's Avatar
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    Re: Functions: Even, Odd or Neither

    Quote Originally Posted by harpazo View Post
    1. F(x) = cuberoot{x}

    F(-x) = cuberoot{-x}

    Result: neither
    Actually, given:

    $\displaystyle f(x)=\sqrt[3]{x}$

    We have:

    $\displaystyle f(-x)=\sqrt[3]{-x}=-\sqrt[3]{x}=-f(x)$

    So, this function is odd.
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    Super Member harpazo's Avatar
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    Re: Functions: Even, Odd or Neither

    Quote Originally Posted by MarkFL View Post
    Actually, given:

    $\displaystyle f(x)=\sqrt[3]{x}$

    We have:

    $\displaystyle f(-x)=\sqrt[3]{-x}=-\sqrt[3]{x}=-f(x)$

    So, this function is odd.
    I did not know that negative 1 can be pulled out of the radicand.
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    Super Member harpazo's Avatar
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    Re: Functions: Even, Odd or Neither

    Quote Originally Posted by topsquark View Post
    Careful! |-x| is not equal, in general, to x.

    $\displaystyle |x| = \begin{cases} -x & x < 0 \\ x & 0 < x \end{cases}$

    We don't know if x is negative or positive. So the only thing you can say is
    $\displaystyle F(-x) = \dfrac{2(-x)}{|-x|} = \dfrac{-2x}{|x|} = -\dfrac{2x}{|x|} = -F(x)$

    As HallsofIvy and Plato have already told you.

    You have a similar problem with #3.


    -Dan
    So, the answer to question 2 is odd.
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    Super Member harpazo's Avatar
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    Re: Functions: Even, Odd or Neither

    Question 3


    f(x) = x + |x|

    f(-x) = -x + |-x|

    f(-x) = -x + |x|

    The function is odd.
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    Re: Functions: Even, Odd or Neither

    Quote Originally Posted by harpazo View Post
    Question 3
    $f(x) = x + |x|$
    The function is odd.
    $-f(x) = -(x + |x|)=-x-|x|\ne -x+|-x|=f(-x)$ So is it odd?
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    Re: Functions: Even, Odd or Neither

    Quote Originally Posted by harpazo View Post
    I did not know that negative 1 can be pulled out of the radicand.
    \begin{align} y=\sqrt[3]{x} \implies y^3 &= x \\
    \text{so}\;{-x} &= -y^3 = (-1)y^3 \\ &= (-1)^3y^3 = \big((-1)y\big)^3 = (-y)^3 \\
    \text{thus}\; \sqrt[3]{-x} &= -y = -\sqrt[3]{x} \end{align}
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  14. #14
    Forum Admin topsquark's Avatar
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    Re: Functions: Even, Odd or Neither

    Quote Originally Posted by harpazo View Post
    I did not know that negative 1 can be pulled out of the radicand.
    Note: $\displaystyle \sqrt[n]{x} = (x)^{1/n}$.

    $\displaystyle \sqrt[n]{-x} = - \sqrt[n]{x}$ only if n is an odd number.

    -Dan
    Last edited by topsquark; Dec 16th 2018 at 07:18 PM. Reason: Made an oopsie
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    Super Member harpazo's Avatar
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    Re: Functions: Even, Odd or Neither

    Quote Originally Posted by Plato View Post
    $-f(x) = -(x + |x|)=-x-|x|\ne -x+|-x|=f(-x)$ So is it odd?
    It's neither.
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