Results 1 to 4 of 4

Math Help - cubic- proof

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    2

    Talking cubic- proof

    How i can proof that third degree polynomial has a point of inflection.

    Grettings. Ignis
    Last edited by ignis; September 7th 2010 at 12:49 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by ignis View Post
    How i can proof that third degree polynomial has a point of inflection.

    Grettings. Ignis
    Show that the derivative has a turning point.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,686
    Thanks
    617
    Hello, ignis!

    How i can prove that third-degree polynomial has a point of inflection?

    I assume you mean a third-degree polynomial function.


    The general cubic function is: . f(x) \:=\:ax^3 + bx^2 + cx +d,\;\;a \ne 0

    The first derivative is: . f'(x) \:=\:3ax^2 + 2bx + c

    The second derivative is: . f''(x) \:=\:6ax + 2b


    A point of inflection occurs where f''(x) \,=\,0

    . . 6ax + 2b \:=\:0 \quad\Rightarrow\quad x \:=\:-\dfrac{b}{3a}

    Therefore, there is an inflection point at: . \left(-\dfrac{b}{3a},\:\dfrac{2b^3-9abc + 27a^2d}{27a^2}\right)
    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 Soroban View Post
    Hello, ignis!


    I assume you mean a third-degree polynomial function.


    The general cubic function is: . f(x) \:=\:ax^3 + bx^2 + cx +d,\;\;a \ne 0

    The first derivative is: . f'(x) \:=\:3ax^2 + 2bx + c

    The second derivative is: . f''(x) \:=\:6ax + 2b


    A point of inflection occurs where f''(x) \,=\,0

    . . 6ax + 2b \:=\:0 \quad\Rightarrow\quad x \:=\:-\dfrac{b}{3a}

    Therefore, there is an inflection point at: . \left(-\dfrac{b}{3a},\:\dfrac{2b^3-9abc + 27a^2d}{27a^2}\right)
    It should be noted that the condition f''(x) = 0 is a necessary but not sufficient condition. f(x) will only have a point of inflection if f''(x) = 0 AND f'(x) has a turning point at the value of x given by the solution to f''(x) = 0.

    Consideration of the graph of f(x) = x^4 illustrates why ....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Lost on Cubic Epsilon Delta proof
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 26th 2010, 08:00 PM
  2. cubic equation proof needed
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 10th 2010, 06:54 AM
  3. Consider the cubic y = f(x) = x^2(ax-b)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 28th 2010, 09:17 AM
  4. Cubic
    Posted in the Algebra Forum
    Replies: 7
    Last Post: October 18th 2008, 06:49 PM
  5. Cubic
    Posted in the Math Topics Forum
    Replies: 6
    Last Post: February 13th 2006, 02:05 AM

Search Tags


/mathhelpforum @mathhelpforum