Results 1 to 5 of 5

Math Help - cubic equation

  1. #1
    Senior Member furor celtica's Avatar
    Joined
    May 2009
    Posts
    271

    cubic equation

    the equation 3x^2 - x^3 = k has three real solutions for k
    write down the set of possible values for k

    i dont understand: can't k = anything as long as x remains real? and what is meant by '3x^2 - x^3 = k has three real solutions for k'?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    3x^2-x^3=k\Leftrightarrow x^3-3x^2+k=0

    Let f(x)=x^3-3x^2+k, \ f:\mathbb{R}\to\mathbb{R}

    f'(x)=3x^2-6x

    f'(x)=0\Rightarrow x_1=0, \ x_2=2

    f(0)=k, \ f(2)=k-4, \ \lim_{x\to-\infty}f(x)=-\infty, \ \lim_{x\to\infty}f(x)=\infty

    The Rolle's sequence is: -\infty \ \  k \ \  k-4 \ \  \infty

    To have 3 real solutions the following inequalities must hold:

    \left\{\begin{array}{ll}k\geq 0\\k-4\leq 0\end{array}\right.\Rightarrow k\in[0,4]
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2009
    Posts
    28
    Draw the graph of the left hand side of the equation. Where you can draw a horizontal line such that the line intersects the graph three times is a value of k that you want - find all such values.

    (drawing is meant for illustration only - it will need to be worked out mathematically. hint: consider turning points)
    Last edited by bananaxxx; July 20th 2009 at 07:59 AM. Reason: added hint
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2009
    Posts
    28
    Quote Originally Posted by red_dog View Post
    3x^2-x^3=k\Leftrightarrow x^3-3x^2+k=0

    Let f(x)=x^3-3x^2+k, \ f:\mathbb{R}\to\mathbb{R}

    f'(x)=3x^2-6x

    f'(x)=0\Rightarrow x_1=0, \ x_2=2

    f(0)=k, \ f(2)=k-4, \ \lim_{x\to-\infty}f(x)=-\infty, \ \lim_{x\to\infty}f(x)=\infty

    The Rolle's sequence is: -\infty \ \  k \ \  k-4 \ \  \infty

    To have 3 real solutions the following inequalities must hold:

    \left\{\begin{array}{ll}k\geq 0\\k-4\leq 0\end{array}\right.\Rightarrow k\in[0,4]
    or red_dog will just give you the answer...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member furor celtica's Avatar
    Joined
    May 2009
    Posts
    271
    thank you both
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cubic equation
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: October 19th 2011, 03:28 PM
  2. Cubic Equation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 15th 2009, 05:16 PM
  3. Help! Cubic equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 19th 2008, 12:02 AM
  4. Cubic Equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 21st 2008, 11:45 PM
  5. Cubic Equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 23rd 2008, 12:00 AM

Search Tags


/mathhelpforum @mathhelpforum