1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rolles Theorem/ Mean Value Theorem + First Derivative Test

  1. Mar 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose that f(x) is a twice-differentiable function defined on the closed interval [a,b]. If f'(c) = 0 for a < c < b, which of the following must be true?

    I. f(a) = f(b)

    II. f has a relative extremum at x = c.

    III. f has a point of inflection at x = c.

    2. Relevant equations

    Rolles Theorem states that there is a c such that f'(c) = 0 between [a,b] if f(x) is continuous on [a,b] and differentiable on (a,b).


    3. The attempt at a solution

    By Rolles Theorem, statement "I" must be correct. (That is, the endpoints are equal to each other). Is this correct?

    I also put statement "II" as correct because since f'(c) = 0 for a < c < b... c must be a critical point or relative extremum. Question though: does this mean there are other points between a and b where f'(c) = 0 ?

    For statement "III", I said this was incorrect. It's twice differentiable yes, but we don't know if f(x) has a point of inflection at x = c.

    Thanks!
     
  2. jcsd
  3. Mar 27, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    You said: "if f'(c) = 0 for a < c < b."
    Did you mean for all c between a and b, or for some c between a and b?

    Because in the first case, for example I is true (although I think it follows from the principal theorem of analysis) while in the second case it is not (you can easily think of a counter example for some specific a, b, c and function f).
     
  4. Mar 27, 2009 #3
    I think there is one more condition here that f(a) = f(b)
    Otherwise there is the counter example of f(x)=ex
     
  5. Mar 27, 2009 #4
    Thank you both. aniketp, you're right. f(a) = f(b) is the last condition... I overlooked it... and since a < b, they can't be equal to each other.
     
  6. Mar 28, 2009 #5

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Are you saying that if a < b then f(a) cannot be equal to f(b) ?
     
  7. Mar 28, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    As for "I", you are confusing the theorem with its converse. Rolle's theorem says that if f(a)= f(b) (and other conditions) then there exist c such that f'(c)= 0. Saying that some f'(c)= 0 does NOT means that f(a) must equal f(b).

    Consider [itex]f(x)= x^2[/itex], a= -1, b= 2, c= 0. f'(0)= 0 but [itex]f(-1)\ne f(2)[/itex]. Or [itex]g(x)= x^3[/itex], a= -1, b= 2. Again g'(0)= 0 but [itex]g(-1)\ne g(2)[/itex].

    f has a relative extremum at 0 but not an inflection point. g has an inflection point at 0 but not a relative extremum. I would say none of those are necessarily true.
     
  8. Mar 28, 2009 #7
    Thank you all! And yes... I did confuse Rolle's Theorem. Thanks for the heads up
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rolles Theorem/ Mean Value Theorem + First Derivative Test
  1. Sylow's First Theorem (Replies: 2)

Loading...