Results 1 to 3 of 3

Thread: Mean Value Theorem

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    1

    Mean Value Theorem

    my problem I am working on is f(x)= x+(54/x); [6,9]

    First I took the first derivative and found that it was f ' (x)= 1-(54/x^2)

    I then found f(6) = 15 and f(9) = 15

    If I plug those into the formula f(b)-f(a) I get 0 this just doesn't make
    b-a 3


    This just doesn't make sense!! Could someone help me?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,410
    Thanks
    1
    What doesn't make sense?

    $\displaystyle f'(c) = \frac{f(b)- f(a)}{b-a} \: \: c \in (a,b)$

    Now, you found that f'(c) = 0, i.e. there exists some point on your interval such that its derivative is equal to 0. What does it exactly mean for your function to have its derivative equal to 0 at c?

    Hint: Rolle's Theorem, a special case of the mean value theorem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by kelleannmoore@yahoo.com View Post
    my problem I am working on is f(x)= x+(54/x); [6,9]

    First I took the first derivative and found that it was f ' (x)= 1-(54/x^2)

    I then found f(6) = 15 and f(9) = 15

    If I plug those into the formula f(b)-f(a) I get 0 this just doesn't make
    b-a 3


    This just doesn't make sense!! Could someone help me?
    The Mean Value Theorem states that if $\displaystyle f(x)$ is continous on some interval $\displaystyle D=[a,b]$ and differentiable on $\displaystyle (a,b)$ then

    $\displaystyle \exists{c}\in{D}\backepsilon~f'(c)=\frac{f(b)-f(a)}{b-a}$

    So we see that we have a special case of the Mean Value Theroem called Rolle's Theorem that states, that if our function meets the same conditions as before and that $\displaystyle f(a)=f(b)$

    Then $\displaystyle \exists{c}\in{D}\backepsilon~f'(c)=0$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Jan 10th 2011, 08:51 AM
  2. Replies: 3
    Last Post: May 14th 2010, 10:04 PM
  3. Prove Wilson's theorem by Lagrange's theorem
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Apr 10th 2010, 01:07 PM
  4. Replies: 2
    Last Post: Apr 3rd 2010, 04:41 PM
  5. Replies: 0
    Last Post: Nov 13th 2009, 05:41 AM

Search Tags


/mathhelpforum @mathhelpforum