Results 1 to 5 of 5

Math Help - Locating Concavity

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    81

    Locating Concavity

    <br />
\displaystyle f(x) = x^{4}-3x^{3}+3x^{2}<br />

    Find the open intervals for which concavity is open upwards.

    <br />
\displaystyle f'(x)=4x^3-9x^2+6x<br />

    <br />
\displaystyle f''(x)=12x^2-18x-6<br />

    I set f''(x)=0
    <br /> <br />
\displaystyle 6(2x^2-3x-1)=0<br />

    <br />
\displaystyle \frac{3\pm\sqrt{17}}{4}<br />

    So when I test the points to the left & to the right, and in-between, the positive intervals are:
    <br />
\displaystyle (-\infty,\frac{3-\sqrt{17}}{4}],[\frac{3+\sqrt{17}}{4},\infty)<br />

    which is apparently wrong. Where did I mess up? Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Well, one mistake is in taking the second derivative: the derivative of +6x is not -6.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    Posts
    660
    Thanks
    133
    Concave up if the second derivative is positive, concave down if the second derivative is negative. If the second derivative is zero, your usually (but not always) at a point where the concavity is changing from up to down or from down to up. (An example of the not always case is y=x^{4} at the origin.)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2010
    Posts
    81
    Got it. Thanks.

    However, this one is question has got me stumped.

    <br />
f(x) = -x^{4}\ln\!\left(x\right)<br />

    I need to find the intervals where f is increasing and decreasing.

    So I start by taking the first derivative:
    <br />
f'(x)=-3x^3lnx-x^3<br />
    <br />
-x^3(3lnx+1)=0<br />
    <br />
x=0 , e^{\frac{-1}{3}}<br />
    since 0 is not in the natural domain, it is not a critical point right?

    When I plug in points on a number line, it shows that my function is increasing on:
    (0,e^{\frac{-1}{3}}] and decreasing on [e^{\frac{-1}{3}},\infty)

    This is wrong. Where did I mess up?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2009
    Posts
    660
    Thanks
    133
    You made a mistake with the differentiation again !
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Locating absolute extrema
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 7th 2011, 05:31 PM
  2. Locating stationary points on a curve
    Posted in the Calculus Forum
    Replies: 7
    Last Post: July 21st 2010, 09:21 AM
  3. Locating the mirror and detector
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 24th 2010, 10:21 PM
  4. Pre Calculus/Trig Locating Lost Treasure Project
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: February 10th 2008, 07:52 PM
  5. Locating periodic solutions (dynamic systems)
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: July 6th 2006, 11:21 AM

Search Tags


/mathhelpforum @mathhelpforum