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Math Help - World of Functions

  1. #1
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    World of Functions

    Determine Algebraically if the functions are even, odd or neither.

    (1) F(x) = cuberoot{x}...I said neither. Is this correct?

    (2) F(x) = 2x/|x|...I said odd. Is this correct?
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  2. #2
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    Quote Originally Posted by magentarita View Post
    Determine Algebraically if the functions are even, odd or neither.

    (1) F(x) = cuberoot{x}...I said neither. Is this correct?

    (2) F(x) = 2x/|x|...I said odd. Is this correct?
    You are correct about (2) but (1) is odd.

    Because:

     f(x) = x^{\frac{1}{3}}

     -f(-x) = -1(-x)^{\frac{1}{3}}
     = -1((-1)^{\frac{1}{3}}(x)^{\frac{1}{3}})
    = ((-1)^3(-1))^{\frac{1}{3}}x^{\frac{1}{3}}
    =((-1)(-1))^{\frac{1}{3}}x^{\frac{1}{3}}
    = 1^{\frac{1}{3}}x^{\frac{1}{3}}
    =x^{\frac{1}{3}}

    By definition, a function is odd if  f(x) = -f(-x)
    Last edited by Mush; January 5th 2009 at 10:56 PM.
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  3. #3
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    ok...

    Quote Originally Posted by Mush View Post
    You are correct about (2) but (1) is odd.

    Because:

     f(x) = x^{\frac{1}{3}}

     -f(-x) = -1(-x)^{\frac{1}{3}}
     = -1((-1)^{\frac{1}{3}}(x)^{\frac{1}{3}})
    = ((-1)^3(-1))^{\frac{1}{3}}x^{\frac{1}{3}}
    =((-1)(-1))^{\frac{1}{3}}x^{\frac{1}{3}}
    = 1^{\frac{1}{3}}x^{\frac{1}{3}}
    =x^{\frac{1}{3}}

    By definition, a function is odd if  f(x) = -f(-x)
    I forgot to examine the cuberoot of {x} as x raised to the 1/3. Interesting reply. Thanks.
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