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Math Help - Families of functions

  1. #1
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    Families of functions

    When is a function neither odd or even?For example why is x^3/2 neither odd or even?

    When the domain is R\{0}, is the range always R+?

    How do u find {x:x^3/2>x^2 } and {x:x^-3/2<x^-2}?
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    Quote Originally Posted by chaneliman View Post
    When is a function neither odd or even?For example why is x^3/2 neither odd or even?
    \frac {x^3}2 is odd. did you mean to say x^{3/2} ?

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

    a function is odd if f(-x) = -f(x)

    otherwise, it is neither

    When the domain is R\{0}, is the range always R+?
    consider the function f(x) = 1/x

    How do u find {x:x^3/2>x^2 } and {x:x^-3/2<x^-2}?
    set up your inequalities and solve for x

    can you solve x^{3/2} > x^2 for x? (is the power 3/2? type clearly)
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    yea i typed it right
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    Quote Originally Posted by chaneliman View Post
    yea i typed it right
    i do not know what you are refering to. you should quote the part of my message that you are responding to so i can respond
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    i can't solve x^3/2>x^2 for x. Does it have something 2 do with logs
    ?
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    Quote Originally Posted by chaneliman View Post
    i can't solve x^3/2>x^2 for x. Does it have something 2 do with logs
    ?
    x^{3/2} > x^2

    Divide by x^{3/2}

    1 > \sqrt{x}

    0 < x < 1
    Last edited by colby2152; January 3rd 2008 at 12:00 PM.
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    Quote Originally Posted by colby2152 View Post
    x^{3/2} > x^2

    Divide by x^{3/2}

    1 > \sqrt{x}

    0 \le x < 1
    Before you divide by x^{3/2} you should say that x^{3/2} > 0 because otherwise the inequality is flipped.
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    Quote Originally Posted by ThePerfectHacker View Post
    Before you divide by x^{3/2} you should say that x^{3/2} > 0 because otherwise the inequality is flipped.
    True, but I kept that as an unstated assumption since it holds in the end o fthe solution with the square root.
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    Quote Originally Posted by colby2152 View Post
    [tex]
    0 \le x < 1
    another technical note, which i'm sure was just a typo or something on your part. we can't have x = 0, since dividing by x^{3/2} would be invalid, and also the original inequality would not hold in the first place. we need x > 0 here for it to make sense
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    Quote Originally Posted by Jhevon View Post
    another technical note, which i'm sure was just a typo or something on your part. we can't have x = 0, since dividing by x^{3/2} would be invalid, and also the original inequality would not hold in the first place. we need x > 0 here for it to make sense
    I didn't see that at last glance, but I originally had a regular greater than sign, but changed it for some reason.
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    i understand it till the point u got 0<x<1
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    Quote Originally Posted by chaneliman View Post
    i understand it till the point u got 0<x<1
    we have 1 > \sqrt{x}

    note that the domain of the square root function is x \ge 0, so we must have that for \sqrt{x} to make sense. however, we drop the x = 0 part, because that does not work in the original inequality.

    now, if we square both sides, we get: 1 > x

    so we have 0< x and x < 1, hence 0 < x < 1
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  13. #13
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    then shouldn't it be 0>x if the domain of the square root function is more then or equal to 0
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    Quote Originally Posted by chaneliman View Post
    then shouldn't it be 0>x if the domain of the square root function is more then or equal to 0
    yes. more than or equal to zero is x \ge 0. we reject the = (for the afore mentioned reason) and make it x > 0. we cannot have x < 0, because the square root is not defined for such x's
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