Results 1 to 7 of 7

Math Help - Show that polynomial has one real root

  1. #1
    Senior Member
    Joined
    Apr 2010
    Posts
    487

    Show that polynomial has one real root

    The question:
    Show that the polynomial p, where p(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} has at least one real root.

    I'm not sure what to do. Solving for 0 is difficult with a cubic, so I don't think this is the approach my text is looking for. This question is within some maxima-minima problems, so I'm thinking I need to use that train of thought. As far as I can tell, the derivative of p has no real roots, so I'm still stuck.

    Any suggestions? Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,506
    Thanks
    1403
    The simplest way would be to substitute some values for x and show that you get both positive and negative values for f(x). This means that f(x) must have changed sign somewhere, and thus must have crossed the x axis somewhere.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    Indeed. However, I suspect that we're expected to be more rigorous in our approach.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Glitch View Post
    Indeed. However, I suspect that we're expected to be more rigorous in our approach.
    The suggestion given IS rigorous. Review the Intermediate Value Theorem.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    I know the IVT. It just didn't cross my mind that it would support Prove It's solution. Thanks guys.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Glitch View Post
    The question:
    Show that the polynomial p, where p(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} has at least one real root.

    I'm not sure what to do. Solving for 0 is difficult with a cubic, so I don't think this is the approach my text is looking for. This question is within some maxima-minima problems, so I'm thinking I need to use that train of thought. As far as I can tell, the derivative of p has no real roots, so I'm still stuck.

    Any suggestions? Thanks.
    A polynomial with real coefficients has complex roots that occur in conjugate pairs. Therefore a polynomial of odd order has an odd number of roots but an even number of complex roots and so must have at least one real root.

    CB
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Glitch View Post
    The question:
    Show that the polynomial p, where p(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} has at least one real root.

    I'm not sure what to do. Solving for 0 is difficult with a cubic, so I don't think this is the approach my text is looking for.

    This question is within some maxima-minima problems,

    so I'm thinking I need to use that train of thought. As far as I can tell, the derivative of p has no real roots, so I'm still stuck.

    Any suggestions? Thanks.
    You can find local maxima and minima by equating the derivative to zero.

    \displaystyle\ f(x)=\frac{x^3}{3!}+\frac{x^2}{2!}+x+1

    \displaystyle\ f'(x)=\frac{x^2}{2}+x+1

    The derivative being zero has complex solutions as \displaystyle\ f'(x)=0\Rightarrow\ x=\frac{-1\pm\sqrt{1-2}}{1}

    hence there are no local maxima or minima.

    Also notice that the derivative is always positive, so the graph is constantly increasing.
    To see this, we can examine for which x gives the least value for the derivative


    \displaystyle\ f'(x)=\frac{x^2}{2}+x+1

    f''(x)=x+1

    This is zero, giving least value for the derivative, when x=-1

    \displaystyle\ x=-1\Rightarrow\ f'(x)=\frac{1}{2}

    The slope of the tangent never goes below 0.5 (the graph is always increasing).

    From here, you can again consider Prove It's statements, as f(0)=1

    so the graph crosses the x-axis only once to the left of that.
    This helps show that the polynomial has only one real root,
    in keeping with your title.

    In the body of your question, you mention at least one real root.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probability of at least one real root
    Posted in the Statistics Forum
    Replies: 3
    Last Post: April 19th 2011, 09:16 AM
  2. Replies: 4
    Last Post: November 3rd 2009, 11:33 AM
  3. show equation has exactly one real root
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 10th 2008, 08:29 PM
  4. Replies: 1
    Last Post: March 29th 2008, 11:11 AM
  5. the only real root of the equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 15th 2007, 06:44 PM

Search Tags


/mathhelpforum @mathhelpforum