Results 1 to 3 of 3

Math Help - Calculus Help (deciphering a function and MVT)

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    16

    Calculus Help (deciphering a function and MVT)

    This is the problem:

    Let f be a function defined by f(x)= x3+ax2+bx+c and having the following properties:

    I) The graph of f has a point of inflection at (0, -2).
    II) The average (mean) value of f(x) on the closed interval [0,2] is -3

    a) What are the values of a,b, and c?
    b) Determine the value of x that satisfies the conclusion of the Mean Value Theorem for f on the closed interval [0,3]

    I got that a=0 by calculating the second derivative, and then plugging in the inflection point. But I would really appreciate some help with the rest of the problem. Thanks to anyone who can help me out!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,356
    Thanks
    36
    Quote Originally Posted by alakaboom1 View Post
    This is the problem:

    Let f be a function defined by f(x)= x3+ax2+bx+c and having the following properties:

    I) The graph of f has a point of inflection at (0, -2).
    II) The average (mean) value of f(x) on the closed interval [0,2] is -3

    a) What are the values of a,b, and c?
    b) Determine the value of x that satisfies the conclusion of the Mean Value Theorem for f on the closed interval [0,3]

    I got that a=0 by calculating the second derivative, and then plugging in the inflection point. But I would really appreciate some help with the rest of the problem. Thanks to anyone who can help me out!!
    You also know that the inflection point is located at (0,-2) so

    f(0) = 0^3 + b\cdot 0 + c = -2 so this give you c=-2.

    Nest, the average of the function is -3

    so

    \frac{1}{2} \int_0^2 \left( x^3 + bx -2 \right)\, dx = -3

    which gives b and thus f(x). For the last part find x_0 such that

    f'\left( x_0 \right) = \frac{f(3) - f(0)}{3}.

    I think you can handle it from here.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2008
    Posts
    16
    Quote Originally Posted by danny arrigo View Post
    You also know that the inflection point is located at (0,-2) so

    f(0) = 0^3 + b\cdot 0 + c = -2 so this give you c=-2.

    Nest, the average of the function is -3

    so

    \frac{1}{2} \int_0^2 \left( x^3 + bx -2 \right)\, dx = -3

    which gives b and thus f(x). For the last part find x_0 such that

    f'\left( x_0 \right) = \frac{f(3) - f(0)}{3}.

    I think you can handle it from here.
    Yes, thank you very much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help deciphering a function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 1st 2008, 04:58 AM
  2. Help with deciphering a function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 30th 2008, 08:31 AM
  3. calculus of modulus function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 13th 2008, 09:47 PM
  4. function calculus expression
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 18th 2008, 06:30 PM
  5. 3 Function/Pre-Calculus Questions
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: October 28th 2006, 02:37 PM

Search Tags


/mathhelpforum @mathhelpforum