Results 1 to 6 of 6

Math Help - Calculus Graph

  1. #1
    NeF
    NeF is offline
    Newbie
    Joined
    Jul 2006
    From
    Cape Town, RSA
    Posts
    17

    Question Calculus Graph


    Clicky Clicky

    I am unable to determine the co-ord of C... well I get a value but it doesnt look right, I get  (1 ; 0) which does not seem to fit the value of C. All the other values seem to be perfect execpt for C. Here are my workings.

    14.1)  f(x) = (x-2)^2(2x-1)=0
            (x^2-4x+4)(2x-1)=0
             2x^3-9x^2+12x-4=0
    A = y-intercept
     y=-4 (0 ; -4)


    B = (2x-1)
     x = 1/2
    (1/2 ; 0) =  B


     C = f(x) = 2x^3-9x^2+12-4
          f'(x) = 6x^2-18+12<br />
                         x^2-3x+2=0
                   (x-1)(x-2)=0
          f(1) = 6(1)^2-18(1) +12 = 0
          C=( 1; 0)

         f(2) = 6(2)^2-18(2)+12=0
                  (2;0) = D Which can also be taken from (x-2)

    Either the book is incorrect or I am. The book has been incorrect a few times so ye :/
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    To find C you note that it is a turning point.
    Thus,
    f'(x)=0
    Since,
    f(x)=(x-2)^2(2x-1)
    The derivative is, (product rule)
    f'(x)=((x-2)^2)'(2x-1)+(x-2)^2(2x-1)'
    Use the chain rule now,
    f'(x)=2(x-2)(2x-1)+2(x-2)^2
    Factor,
    f'(x)=2(x-2)[(2x-1)+(x-2)]
    Simplfy,
    f'(x)=2(x-2)(3x-3)
    You can factor again,
    f'(x)=6(x-2)(x-1)
    Set equal to zero,
    6(x-2)(x-1)=0
    Divide by 6,
    (x-2)(x-1)=0
    Each factor equal to zero,
    x-2=0\rightarrow x=2
    x-1=0\rightarrow x=1
    One of these points is C and one is D, but which one is it?
    Note C is to the left of D thus C<D.
    Thus,
    x=1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    NeF
    NeF is offline
    Newbie
    Joined
    Jul 2006
    From
    Cape Town, RSA
    Posts
    17
    Thanks. I havent seen that way of solving it like that. I agree that C is 1 but look where it is on the graph. But how can C be (1;0) look where it is on the grapg. That pos means that it has a x and y value. I am confused about that
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    I would also like to know how to solve for y PH
    do you just substitute x=1 into the original equation?
    Attached Thumbnails Attached Thumbnails Calculus Graph-calculus-graph.jpg  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Quick
    I would also like to know how to solve for y PH
    do you just substitute x=1 into the original equation?
    Yes,
    so the point is
    (1,1), not what you said NeF, (1,0) that is wrong.
    ----
    I think what your problem is. You are thinking since it is a turning point it is zero. Wrong! The derivative is zero, not necessarily the point is zero.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    NeF
    NeF is offline
    Newbie
    Joined
    Jul 2006
    From
    Cape Town, RSA
    Posts
    17
    thanx for clearing that up
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Graph Theory / Chromatic Number of a Complete Graph
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 15th 2011, 10:59 AM
  2. Replies: 0
    Last Post: September 25th 2010, 06:59 AM
  3. Replies: 1
    Last Post: February 15th 2009, 06:35 AM
  4. Replies: 1
    Last Post: September 8th 2008, 06:31 AM
  5. calculus graph, please check, thanx
    Posted in the Calculus Forum
    Replies: 9
    Last Post: July 25th 2006, 10:29 PM

Search Tags


/mathhelpforum @mathhelpforum