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I'm going to prove the M.V.T. and use it to prove Rolle's Theorem.

Statement of Mean Value Theorem

  • A function f is continuous on a closed interval [a,b] and differentiable on every point in the interval (a,b) such that for a < c (some point on the closed interval) < b and f(b)-f(a)/b-a = f'(c).

I will show that there exists a point c within the interval where f'(c)=0.

Let g(x)=(f(b)-f(a))x-(b-a)f(x) where x is substituted for c.
Note that g'(x)=(f(b)-f(a))-(b-a)f'(x)

Setting the derivative equal to zero leaves f'(x)=f(b)-f(a)/b-a.
Let a=x and g=f such that:
g(a)=(f(b)-f(a))a-(b-a)f(a).
g(a)= a*f(b)-a*f(a)-b*f(a)+a*f(a)
g(a)=a*f(b)-b*f(a)
And
g(b)=(f(b)-f(a))b-(b-a)f(b)
g(b)=b*f(b)-b*f(a)-b*f(b)+a*f(b) g(b)=a*f(b)-b*f(a)
We can conclude that g(a)=g(b).

  • If g'(x)=0 on some point in the closed interval, then g'(x) is zero for every point on the interval.
  • If g(a) > g(b) on some point x in the closed interval, then c is a point that is a local maximum and by definition the derivative at a local maximum/minimum is 0 concluding that g'(x)=0.
  • If g(a) < g(b) on some point x in the closed interval where c is a point at which the function g(x) has a local minimum and g'(x)=0. By going through all three cases (constant, maximum, and minimum) there will always be some point c where f'(c)=0.

Recall that g(a)=g(b). Using this to satisfy the third condition of Rolle's theorem that:
f(b)-f(a)/b-a=f'(c) will be f'(c)=0 due to f(b)=f(a).

How does it look?

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I believe you are a bit confused. Try to imagine functions like $f(x)=1$ (for all $x$), $f(x)=x$ (for all $x$), $f(x)=x^2$ (for all $x$), etc. These are all differentiable and (hence) continuous on the whole real line, in particular you can apply the MVT on every interval. Now given an interval $I$ is it always true that there is $c\in I $ such that $f(c)=0$? Try to understand the statement for specific functions first - What does the theorem mean geographically? –  AD. Nov 12 '12 at 22:17
    
Usually one proves the special case (Rolle) first, and then use that to prove MVT. –  André Nicolas Nov 12 '12 at 22:24
    
I agree with André. It is much easier to prove MVT using Rolle's. –  glebovg Nov 12 '12 at 23:49
    
André is right about the usual order. What glebovg wrote is ambiguous. I hope (s)he meant that it's easier to prove MVT with Rolle than without; that's surely true. The other possible reading, that it's easier to prove MVT from Rolle than to prove Rolle from MVT (which the OP wants to do), is false. Proving Rolle from MVT is trivial (provided you include the hypothesis of Rolle in your formulation, which the OP apparently didn't do). –  Andreas Blass Nov 13 '12 at 0:13

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