Results 1 to 2 of 2

Math Help - Another proof question

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    9

    Another proof question

    Prove there exists a number c on (-4,1), such that g ' (c) = -2 Give an explanation for your proof. I used the mean value theorem and found that it indeed does work. But how do I word the explanation.

    g(x)= x^3 - 3x^2 - 24x +12

    Maybe...
    For the continuous function g, the endpoints of the function on the interval (-4,1) form a secant line. For the function g, there exists a point such that the tangent line at that point is parallel to the secant line...

    what else should I add...

    thanks,
    D.
    Last edited by dboyd435; December 21st 2008 at 07:17 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member Ziaris's Avatar
    Joined
    Dec 2008
    Posts
    26
    Just take the derivative and use the quadratic formula. Quick and simple relative to using the mean value theorem.

    \frac{dy}{dx}=3x^2-6x-24=-2,3x^2-6x-22=0.

    \frac{6\pm \sqrt{36-4(3)(-22)}}{6}=\frac{3\pm 5\sqrt{3}}{3}.

    The only one that is in (-4,1) is \frac{3-5\sqrt{3}}{3}\approx -1.88675. This both proves the existence of the value and gives the value explicitly.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof question
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: January 21st 2010, 07:31 AM
  2. proof question 2,1
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: October 26th 2009, 02:54 PM
  3. a proof question..
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 25th 2008, 08:35 AM
  4. Proof Question
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 19th 2006, 05:54 AM
  5. Proof question
    Posted in the Algebra Forum
    Replies: 4
    Last Post: February 22nd 2006, 02:50 PM

Search Tags


/mathhelpforum @mathhelpforum