Results 1 to 4 of 4

Math Help - Examining a Curve

  1. #1
    Senior Member
    Joined
    Oct 2012
    From
    USA
    Posts
    325
    Thanks
    1

    Examining a Curve

    All of this is wrong. Here is the actual question: "Find the local maximum and minimum values of f using both the First and Second Derivative Tests." Please look thru this to see what is mistaken.

    f(x) = -8x^{3} + 12x^{2} + 8

    f'(x) = -24x^{2} + 24x

    f''(x) = -48x + 24

    Find Critical Numbers:

    f'(x) = -24x^{2} + 24x

    -24x^{2} + 24x = 0

    -24x^{2} = - 24

    x^{2} = 1

    \sqrt{x^{2}} = \sqrt{1}

    x = \pm \sqrt{1}

    x = -1

    x = 1

    Find Min Max

    f(x) = -8x^{3} + 12x^{2} + 8

    f(-1) = -8(-1)^{3} + 12(-1)^{2} + 8 = 28

    f(1) = -8(1)^{3} + 12(1)^{2} + 8 = 12
    Last edited by Jason76; October 29th 2013 at 12:34 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,612
    Thanks
    591

    Re: Examining a Curve

    Hey Jason76.

    There is nothing wrong with your first answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2010
    From
    Oslo
    Posts
    57
    Thanks
    4

    Re: Examining a Curve

    Quote Originally Posted by Jason76 View Post
    All of this is wrong. Here is the actual question: "Find the local maximum and minimum values of f using both the First and Second Derivative Tests." Please look thru this to see what is mistaken.

    f(x) = -8x^{3} + 12x^{2} + 8

    f'(x) = -24x^{2} + 24x

    f''(x) = -48x + 24

    Find Critical Numbers:

    f'(x) = -24x^{2} + 24x

    -24x^{2} + 24x = 0

    -24x^{2} = - 24
    The last line here is missing an x on the right hand side.

    You can also write f'(x) as:

    f'(x) = 24x(1-x)

    This is zero when x=0 or x=1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Oct 2012
    From
    USA
    Posts
    325
    Thanks
    1

    Re: Examining a Curve

    f(x) = -8x^{3} + 12x^{2} + 8

    f'(x) = -24x^{2} + 24x

    f''(x) = -48x + 24

    Find Critical Numbers:

    f'(x) = -24x^{2} + 24x

    -24x^{2} + 24x = 0

    24x(-x + 1) = 0 a different approach

    24x = 0

    x = 0

    (-x + 1) = 0

    -x = -1

    x = 1 and x = 0 (Critical Number)


    Find Min Max

    f(x) = -8x^{3} + 12x^{2} + 8

    f(1) = -8(1)^{3} + 12(1)^{2} + 8 = 12 - Max

    f(0) = -8(0)^{3} + 12(0)^{2} + 8 = 8 - Min
    Last edited by Jason76; October 29th 2013 at 12:17 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Examining minimizing and solutions
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 13th 2009, 07:11 PM
  2. Invertible & examining the image through the unit square
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: December 13th 2009, 07:50 AM
  3. examining a riemann integral
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 23rd 2009, 10:48 AM
  4. Examining a transformation
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: February 14th 2009, 04:29 AM
  5. Replies: 8
    Last Post: October 8th 2007, 04:29 PM

Search Tags


/mathhelpforum @mathhelpforum