Results 1 to 9 of 9

Math Help - What type of function is this?

  1. #1
    Member
    Joined
    May 2008
    Posts
    112

    What type of function is this?

    y = -x^(1/3)

    What type of function is this? Simplified it is y = i*cube root(x)

    I'm trying to find the symmetry using the formula x = -b/2a.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,698
    Thanks
    454
    Quote Originally Posted by mwok View Post
    y = -x^(1/3)

    What type of function is this? Simplified it is y = i*cube root(x)

    I'm trying to find the symmetry using the formula x = -b/2a.
    the function you cite is actually f(x) = -\sqrt[3]{x}

    did you mean y = (-x)^{\frac{1}{3}} ?

    if so, this is the same as f(x) = -\sqrt[3]{x}

    note that (-x)^{\frac{1}{3}} = (-1)^{\frac{1}{3}} \cdot x^{\frac{1}{3}} = -1 \cdot x^{\frac{1}{3}}

    no i is involved ... (-1)^{\frac{1}{2}} = i

    f(x) = -\sqrt[3]{x} is an odd function symmetric to the origin.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    112
    But how did you find the symmetry and odd/even?

    I know that an even/odd power corresponds to a even/odd function, however...the power in this case is a fraction. Without graphing, how can I determined this?

    BTW, what graphing tool did you use to generate that graph?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,698
    Thanks
    454
    Quote Originally Posted by mwok View Post
    But how did you find the symmetry and odd/even?

    I know that an even/odd power corresponds to a even/odd function, however...the power in this case is a fraction. Without graphing, how can I determined this?

    BTW, what graphing tool did you use to generate that graph?

    Thanks.
    f(x) is odd if f(-x) = -f(x)

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

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

    f(-x) = \sqrt[3]{x} is the opposite of f(x) = -\sqrt[3]{x} , therefore f(x) is odd.


    free graph program ...

    Graph
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,661
    Thanks
    1616
    Awards
    1
    Quote Originally Posted by mwok View Post
    But how did you find the symmetry and odd/even?
    I know that an even/odd power corresponds to a even/odd function, however...the power in this case is a fraction. Without graphing, how can I determined this?
    The names odd & even most likely came from exponents.
    But exponents really do not determine much.
    For example: \cos(x) is even and \sin(x) is odd.

    Learn the actual definitions: If f(x)=f(-x) then f is even.
    If -g(x)=g(-x) then g is odd.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2008
    Posts
    112
    Thanks, how do you find the symmetry? I cannot use x = -b/2a because that is for polynomials and the function above is not a polynomial.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,661
    Thanks
    1616
    Awards
    1
    Quote Originally Posted by mwok View Post
    Thanks, how do you find the symmetry?
    First do as I suggested: Learn the basic definitions.
    Then it is simple: odd functions are symmetric about the origin; even functions are symmetric about the y-axis.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    May 2008
    Posts
    112
    So, the answer is simply that the symmetry is about the origin?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,661
    Thanks
    1616
    Awards
    1
    Quote Originally Posted by mwok View Post
    So, the answer is simply that the symmetry is about the origin?
    Well of course, if it is an odd function.
    f(x) =  - x^{\frac{1}<br />
{3}} \; \Rightarrow \;f( - x) =  - \left( { - x} \right)^{\frac{1}<br />
{3}}  = x^{\frac{1}<br />
{3}}  =  - f(x) so it is odd.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Super stuck on Type 1 and Type 2 errors
    Posted in the Statistics Forum
    Replies: 0
    Last Post: January 29th 2010, 12:06 PM
  2. Type I and Type II errors with sigma given...
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: January 20th 2010, 03:09 PM
  3. Type 1/ Type 2 Error Question
    Posted in the Statistics Forum
    Replies: 4
    Last Post: August 3rd 2008, 06:51 AM
  4. Replies: 2
    Last Post: January 31st 2008, 05:05 PM
  5. A special type of function...
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 18th 2006, 04:06 AM

Search Tags


/mathhelpforum @mathhelpforum