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Mean value theorm [HARD]

  1. Jul 25, 2007 #1


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    1. The problem statement, all variables and given/known data
    f is a continues function in [a,b] and has a second derivative in (a,b). L(x) is a line that goes through (a,f(a)) and (b,f(b)).
    Prove that if f''(x)>0 in (a,b) then L(x)>f(x) for every x in (a,b)

    2. Relevant equations


    3. The attempt at a solution

    First of all,
    [tex] L(x) = f(a) + \frac{f(b)-f(a)}{b-a} (x-a) [/tex]
    and so
    [tex] L(x) - f(x) = f(a) - f(x) + \frac{f(b)-f(a)}{b-a} (x-a) [/tex]
    [tex] \frac{f(b)-f(a)}{b-a} = f'(t) [/tex] where t is in (a,b)
    and so
    [tex] ( L - f)' (x) = f'(t) - f'(x) [/tex]
    Now, [tex] ( L - f )' (x) = 0 [/tex] only when [tex] f'(t) = f'(x) [/tex]
    And since f''(x)>0 f'(x) is an injective function in (a,b) and so [tex] f'(t) = f'(x) [/tex] only when x=t. and since f''(x)>0 we get a maximum at x=t.
    Now, (L-f)(a) = 0 and (L-f)(b) = 0. If for any other x_0 =/= t
    (L-f)(x_0)=0 then that would mean that for some x in (a,x_0) and for some x in (x_0,b) (L-f)'(x) = 0, but this is impossible as (L-f)'(x) is injective and we alredy found one point (t) where (L-f)'(x) = 0.
    Also, (L-f)(t) > (L-f)(a) = 0 because it's a maximum in (a,b). And so for all x in (a,b) L-f)(x) > 0 => L(x) > f(x).

    I felt that that proof was pretty weak, especially towards the end. How can I make it better. And is there any easier way?
  2. jcsd
  3. Jul 25, 2007 #2


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    Hint (hopefully): Let f : I --> R be differentiable twice on an open real interval I. f is a convex function on I iff f''(x) >= 0, for every x in I.
  4. Jul 25, 2007 #3
    This is a proof that f is a convex function. As far as I know, the only way to do it is with a contradiction. Suppose there exists a t in (a,b) with [itex]L(t) \le f(t)[/itex]. That ought to make it feel more cleaner and mathy-like.

    Your logic is backwards, you're presupposing (L-f)'(x)=0 for some x. It makes more sense that since (L-f)(a)=0 and (L-f)(b)=0 that there would exist a t in (a,b) such that (L-f)'(t)=0.

    "and since f''(x)>0 we get a maximum at x=t." If g'(x)=0 and g''(x)<0 then g(x) is a local maximum. (L-f)'(t)=0 and f''(x)>0 doesn't tell you 'something gets maximized' (I'm not sure what you're saying gets maximized).

    Anyhow, since (L-f)''(x)=-f(x)<0 (L-f)(t) is a local maximum.
    Last edited: Jul 25, 2007
  5. Jul 25, 2007 #4
    Unfortunately that's not a hint. That's precisely the inference he's been asked to show.
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