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Mean Value Theorm type question

  1. Mar 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Let f : R ---> R be a Continuosly differentiable function such that f(0) = 0 , f(1) = 1 and f'(0) = f'(1) = 0 .
    Show that their exists an S in (0, 1) such that |f''(S)| > 4 .

    2. Relevant equations

    I am unsure if I am even supposed to use the mean value theorm in this problem, but it seems like you have to..


    Let f : [a, b] → R be continuous on the closed interval [a, b], and differentiable on the open interval (a, b). Then there exists some R in (a, b) such that

    f(b) − f(a) = f'(R)(b − a)

    3. The attempt at a solution

    Apply derivatives to the mean value theorm above to get:

    f'(b) − f'(a) = f''(S)(b − a)

    Now, taking the magnitudes, we get

    |f'(b) − f'(a)| < |f''(S)||(b − a)|

    Now computing, we get

    |f'(1) − f'(0)| < |f''(S)||(1 − 0)|

    |0 − 0| < |f''(S)|

    0< |f''(S)|


    0< |f''(S)|

    so indeed, 4< |f''(S)|

    How does this sound?

    I am very unsure of this...I dont think it is rigourous enough
  2. jcsd
  3. Mar 4, 2007 #2


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    Staff Emeritus
    Science Advisor

    You are only given that f is "continuously differentiable". It does not follow from that that f even has a second derivative at any particular point.

    Also, f(b)- f(a) = f'(c)(b-a) is true for specific values of a, b, c, not variables. You can say that, for any x, y in [a, b], f(x)- f(y)= f'(c)(x-y) but then what variable are you going to differentiate with respect to: x or y?

    Finally, you can say that, for any x in [a,b], f(x)- f(b)= f'(c)(x-a) and differentiate that with respect to x but then you get f'(x)= f'(c), not what you give.

    You can use the mean value theorem to say that there exist c in [0, 1] such that f'(c)= f(1)- f(0)= 1.
    Last edited: Mar 4, 2007
  4. Mar 4, 2007 #3
    Thank you HallsofIvy... I had the Mean Value Theorm wrong...

    Let me start again..

    a=0, b=1

    for any x in [a,b], f(x)- f(b)= f'(c)(x-a)


    f(x)- f(1)= f'(c)(x-0)

    and we take the derivative with respect to x

    f'(x)- f'(1)= f''(c)(x-0)

    now we take the magnitude

    |f'(x)- f'(1)| < |f''(c)||x|

    and f'(1)=0

    |f'(x)| < |f''(c)||x|

    and now i don't know where to go from here..

    any help please?
  5. Mar 4, 2007 #4
    I think i know how to solve it...

    I need to find a function such that f(0)=0 and f(1)=1 and f'(0)=f'(1)=1.


    f(x) = 0 for 0 =< x =< 0.25

    f(x) = x for 0.25 < x =<0.75

    f(x) = 1 for 0.75< x =< 1

    but off course, f'(x) wont be conintuously differentiable...

    thus, for

    |f'(x)| < |f''(c)||x|

    choose x=0.75


    for x=0.75, f'(x) = 1

    4/3 < |f''(c)|

    but this function dosn't work in my case...

    anybody know a few functions such that f(0)=0 and f(1)=1 and f'(0)=f'(1)=1 ?
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