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Graph sketching

  1. Aug 3, 2010 #1
    1. The problem statement, all variables and given/known data

    I know for graphs of function f(x)=x^n where n is an odd power, even power or square root have their own pattern but how about


    or f(x)=x^(1/2)

    is that considered odd or even ?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 3, 2010 #2

    Gib Z

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    Homework Helper

    No, 3/2 and 1/2 are neither even or odd. There isn't exactly a simple rule for f(x) = x^(3/2), but it is a famous curve called http://mathworld.wolfram.com/SemicubicalParabola.html" [Broken]

    You can think of f(x) = x^(1/2) as a parabola tipped on its side, and then the bottom half is chopped off so that it's a function.
    Last edited by a moderator: May 4, 2017
  4. Aug 3, 2010 #3
    thanks, i have tried graphing several such graphs with a program and notice something.

    Any function f(x)=x^n , where n is 1/2, 1/3 (the denominator can be any real and the numerator is 1), the graph will look like a one-sided parabola opening to the right

    And if n=3/2, 5/2, 7/3 (any rational numbers aside from case 1)

    the graph will look like a semicubical parabola.
    Last edited by a moderator: May 4, 2017
  5. Aug 3, 2010 #4


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    Err, not quite. The graph of [tex]f(x) = x^{1/3}[/tex] does NOT look like a one-sided parabola opening to the right.

    The function [tex]f(x) = x^{2}[/tex] is a parabola, but it is not one-to-one. If we restrict the domain of f(x) to [0, ∞), then f(x) would be one-to-one and the inverse would be [tex]f^{-1}(x) = x^{1/2} = \sqrt{x}[/tex]. So the graph of [tex]f^{-1}(x)[/tex] would be a half-of-a-parabola laying on its side.

    Now, the function [tex]g(x) = x^{3}[/tex], your basic cubic, IS one-to-one, so we don't need to restrict the domain. It's inverse would be [tex]g^{-1}(x) = x^{1/3} = \sqrt[3]{x}[/tex], and its graph would look like the COMPLETE graph of [tex]g(x) = x^{3}[/tex], but rotated to the side and flipped, for a lack of a better desciption.

  6. Aug 3, 2010 #5
    yeah that's only when the denominator is odd. Thanks.
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