Results 1 to 3 of 3

Math Help - calculus

  1. #1
    Newbie
    Joined
    Aug 2008
    Posts
    2

    calculus

    Hello everyone. I hope all are well. This is my first post and visit of many to come in the future I'm sure. My name is Nav and I am currently a college student that went back to school to get my degree.

    I was hoping someone could help me with a couple of homework problems I have. I did most of my homework, but I couldn't get these last two. I believe the teacher purposely made these last two hard.

    1. Show that every (degree 3) polynomial has exactly one point where its second derivative is zero. Call this point P.

    2. The show if a polynomial has three x-intercepts x1, x2, and x3, then x-coordinate of P is equal to x1+x2+x3/3.


    Any help would be greatly appreciated. I am taking calculus after 10 years so I am having a bit of difficulty, but I am learning and picking it up again. I didn't know where else to go. Thank you for your time. Thank you, thank you, thank you, thank you..lol
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by omniconsum3r View Post
    ...

    1. Show that every (degree 3) polynomial has exactly one point where its second derivative is zero. Call this point P.

    2. The show if a polynomial has three x-intercepts x1, x2, and x3, then x-coordinate of P is equal to (x1+x2+x3)/3.


    Any help would be greatly appreciated. I am taking calculus after 10 years so I am having a bit of difficulty, but I am learning and picking it up again. I didn't know where else to go. Thank you for your time. Thank you, thank you, thank you, thank you..lol
    1. Let
    f(x)=ax^3+bx^2+cx+d be the function of third degree.

    Then the second drivative is:

    f''(x) = 6ax+2b . With f''(x) = 0 you'll get: x = -\frac b{3a}

    Calculate f\left(  -\frac b{3a} \right) = -\frac{bc}{3a} + \frac{2b^3}{27a^2} + d ....... Therefore P\left(  -\frac b{3a}~,~ -\frac{bc}{3a} + \frac{2b^3}{27a^2} + d\right)

    2. If the function has 3 zeros you can write f as a productof the linear factors:

    f(x)=a(x-x_1)(x-x_2)(x-x_3) = a(x^3-x_1x^2-x_2x^2-x_3x^2 + uninteresting\ summand) The uninteresting summand contains x to the power of 1 or 0.

    Then the second derivative is:

    f''(x)=a(6x-2(x_1+x_2+x_3)) With f''(x) = 0 you'll get the given result.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2008
    Posts
    2
    You are awesome dude!!! Thank you man, thank you. I understand what you did which is even more awesome!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 13th 2011, 09:11 PM
  2. Replies: 2
    Last Post: June 25th 2010, 10:41 PM
  3. Replies: 1
    Last Post: February 11th 2010, 07:09 AM
  4. Calculus III But doesn't require Calculus :)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 19th 2009, 04:23 PM
  5. Replies: 1
    Last Post: June 23rd 2008, 09:17 AM

Search Tags


/mathhelpforum @mathhelpforum