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Rolle's Theorem differentiation

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data

    Let f be continuous and differentiable on [a,b], and suppose that f attains its maximum and minimum points c and d, respectively, where c,d belong to [a,b]. Show that f ' (d) = 0

    2. Relevant equations



    3. The attempt at a solution

    I thought about using the Mean Value Theorem which states that:

    f ' (d)=f(b) - f(a) / (b-a)

    but then didn't know how to continue.

    Or maybe use Rolle's Theorem, but i couldn't see how that would show that f ' (d) = 0

    Any help would be great thanks.
     
  2. jcsd
  3. Nov 24, 2008 #2
    Re: differentiation

    this isn't necessarily true
    [itex]
    f(x) = x^2
    [/itex]
    [itex]
    x \in [0,10]
    [/itex]
    is continuous and differentiable, f attains it's maximum at f(10)=100 but f'(10) =20. I think you need c,d to belong to (a,b). This way you know that f(b)< f(d) and f(a)< f(d)
     
  4. Nov 24, 2008 #3
    Re: differentiation

    c,d do belong to [a,b]

    so f(a)<f(d), but f(b)>f(d)

    I don't see how this helps me to show that f ' (d) = 0 thought.
     
  5. Nov 24, 2008 #4

    HallsofIvy

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    Re: differentiation

    No, the mean value theorem doesn't tell you that: it says that (f(b)- f(a))/(b-a)= f'(h) for SOME number h between a and d. It does not follow that h is the "d" where this is minimum.

    Rather, use the basic definition of the derivative.
    [tex]f'(d)= \lim_{h\rightarrow 0} \frac{f(d+h)- f(d)}{h}[/tex]

    If d is a (local) minimum, what can you say about f(d+ h) for any small non-zero h?
     
    Last edited: Nov 24, 2008
  6. Nov 24, 2008 #5
    Re: differentiation

    Is the limit when h tends to 0?....then it would be f(d)-f(d)/h, so f '(d) would equal zero.

    Im not sure what f(d+h) implies though....
     
  7. Nov 24, 2008 #6
    Re: differentiation

    first things first
    [itex]
    c,d \in (a,b)
    [/itex]
    instead of [a,b] if you need f'(x) = 0 at min/max points
    agreed?
     
    Last edited: Nov 24, 2008
  8. Nov 24, 2008 #7
    Re: differentiation

    f(c) is the maximum and f(d) is the minimum.

    What can i say about f(d+h) ?
     
  9. Nov 24, 2008 #8

    gabbagabbahey

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    Re: differentiation

    f(d) is the minimum, sooo f(d+h) is....

    (1) smaller than f(d)
    (2) greater than f(d)
    (3)purple

    ?
     
  10. Nov 24, 2008 #9
    Re: differentiation

    ok....so obviously f(d+h)>f(d)

    i still don't see how this can help me show that f '(d) = 0
     
  11. Nov 24, 2008 #10

    HallsofIvy

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    Re: differentiation

    No, that does not follow. Every derivative becomes "0/0"!
     
  12. Nov 24, 2008 #11

    HallsofIvy

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    Re: differentiation

    So f(d+h)- f(d)> 0 no matter whether h is positive or negative.

    If h> 0, what can you say about (f(d+h)- f(d))/h?

    If h< 0, what can you say about (f(d+h)- f(d))/h?

    And so what must be true of
    [tex]\lim_{h\rightarrow 0} \frac{f(d+h)- f(d)}{h}[/tex]?
     
  13. Nov 24, 2008 #12
    Re: differentiation

    ok, so if h>0, then (f(d+h)- f(d))/h >0

    and if h< 0, then (f(d+h)- f(d))/h <0

    which means that the limit eqn will be zero.
     
  14. Nov 24, 2008 #13

    HallsofIvy

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    Re: differentiation

    WHY does that mean the limit must be 0?
     
  15. Nov 24, 2008 #14
    Re: differentiation

    i dont know, im sorry, i just don't understand where you're going with this...
     
  16. Nov 24, 2008 #15

    gabbagabbahey

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    Re: differentiation

    If [tex]\lim_{x \to 0}[/tex] of some expression exists, what must be true about [tex]\lim_{x \to 0^+}[/tex] and [tex]\lim_{x \to 0^-}[/tex]?

    In this case, you are told that f is differentiable, so you know that the limit [tex]
    \lim_{h\rightarrow 0} \frac{f(d+h)- f(d)}{h}
    [/tex] must exist. So....
     
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