Results 1 to 3 of 3

Math Help - urgent- Analysis differentiation

  1. #1
    C.E
    C.E is offline
    Junior Member
    Joined
    Mar 2009
    Posts
    34

    urgent- Analysis differentiation

    Can somebody please help I do not know how to do this and have an exam tommorow!

    Suppose that f : [a; b] --> R is a differentiable function, and assume f'(a) <
    k < f'(b). Show that there exists c in (a; b) with f'(c) = k. Do NOT assume that f' is
    continuous! Hint: consider g(x) = f(x) - kx.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301
    Quote Originally Posted by C.E View Post
    Can somebody please help I do not know how to do this and have an exam tommorow!

    Suppose that f : [a; b] --> R is a differentiable function, and assume f'(a) <
    k < f'(b). Show that there exists c in (a; b) with f'(c) = k. Do NOT assume that f' is
    continuous! Hint: consider g(x) = f(x) - kx.
    Proof. So  g'(x) = f'(x)-k .  g attains a maximum on  [a,b] by the Extreme Value Theorem. This is achieved at some  c \in (a,b) since  g'(a) >0 and  g'(b) < 0 . By Fermat's Theorem,  g'(c) = 0 . Thus  f'(c) = k .  \square
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    Here's a slightly different proof that, although derivative functions are not necessarily continuous, they satisfy the "intermediate value" property, which is, I believe, what Sampras is referring to as the "Fermat's theorem".

    Define h(x)= (f(x)- f(a))/(x- a). Then h is continuous and, at x= a, h(a)= f'(a). Therefore, h takes on all values between f'(a) and h(b)= (f(b)- f(a))/(b-a) on the interval [a, b]. Further, by the mean value theorem applied to the interval [a, x], there exist c such that f'(c)= (f(x)- f(a))/(x- a). That is, f' takes on all such values.

    Define j(x)= (f(b)- f(x))/(b- x). Then j is continuous and, at x= b, j(b)= f'(b).
    Therefore, j takes on all values between j(a)= (f(b)- f(a))/(b-a) and f'(b). Further, by the mean value theorem applied to the interval [x, b], there exist c such that f'(c)= (f(b)- f(x))/(x- a). That is, f' takes on all such values. Since those two intervals include all numbers between f'(a) and f'(b), for any k such that f'(a)< k< f'(b), there exist c such that f'(c)= k.

    This shows, in turn, by the way, that if \lim_{x\rightarrow a} f'(x) exists, then it must be equal to f'(a). The only way a derivative can not be continuous at a point is if that limit does not exist there.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. VERY hard differentiation analysis questions
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: May 31st 2009, 07:08 AM
  2. Analysis differentiation
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 30th 2009, 03:36 PM
  3. [Urgent] help!! calculus, function analysis.
    Posted in the Calculus Forum
    Replies: 8
    Last Post: December 3rd 2008, 07:16 PM
  4. Differentiation (Real Analysis)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 30th 2008, 05:32 PM
  5. Differentiation Analysis
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 26th 2007, 06:01 PM

Search Tags


/mathhelpforum @mathhelpforum