prove that suppose 'f' is continious on [a,b], f'(x) exists for each x
in the (a, b) , then there is a c in (a,b) suchthat
f'(c)=(f(b)-f(a))/(b-a).

Hi,

Thank you for your question.

The proof of the mean value theorem requires Rolle's theorem.

Rolle's theorem states:
Let f be differentiable on (a,b) and continuous on [a,b]. If f(a) =
f(b) = 0, then there is at least one point c in (a,b) for which f'(c)
= 0.

Note that either f(x) is always 0 on [a,b] or f varies on [a,b]. 

Case 1: If f is not positive or negative on the interval [a,b], then f
must be zero everywhere on the interval. That is, f = 0, and in
particular, f is constant. The derivative of a constant function is
zero, so f'(c) = 0 for all c between a and b. Thus, the conclusion of
Rolle's theorem holds in this case.

Case 2: f is positive somewhere on the interval [a,b]
Suppose that f is positive somewhere on the interval [a,b]. A
continuous function on a closed interval takes on a maximum value
somewhere on the interval; suppose that f attains its max at x = c.
Since I know f is positive somewhere on the interval, the max f(c)
must be positive; since f(a) = 0 and f(b) = 0, c can't be either a or
b. Therefore, a < c < b. In particular, f is differentiable at c. But
at a point where a differentiable function attains a max, the
derivative is zero. Thus, f'(c) = 0, and Rolle's theorem holds in this
case as well.

Case 3: f is negative somewhere on the interval [a,b]
Suppose that f is negative somewhere on the interval [a,b]. A
continuous function on a closed interval takes on a minimum value
somewhere on the interval; suppose that f attains its min at x = c.
Since I know f is negative somewhere on the interval, the min f(c)
must be negative; since f(a) = 0 and f(b) = 0, c can't be either a or
b. Therefore, a < c < b. In particular, f is differentiable at c. But
at a point where a differentiable function attains a min, the
derivative is zero. Thus, f'(c) = 0, and Rolle's theorem holds in Case
3.

That proves Rolle's theorem. This proof is from:

A Proof of Rolle's Theorem
http://www.millersv.edu/~bikenaga/calculus/mvt/rollepf.html



We can now prove the mean value theorem.

Let us write the equation for the secant through (a,f(a)) and (b,f(b)):

          f(b)-f(a)
y-f(a) =  ---------   (x-a) 
            b-a
 

which we can rewrite as 
     f(b)-f(a)
y = ----------  (x-a)+f(a). 
       b-a
 

Let 

               f(b)-f(a)
g(x) = f(x)-[ ----------  (x-a)+f(a) ]
                  b-a


Note that g(a) = g(b) = 0. Also, g is continuous on [a,b] and
differentiable on (a,b) since f is. So by Rolle's Theorem there exists
c in (a,b) such that
g'(c)= 0. 

But g'(x) = f'(x)-[(f(b)-f(a))/( b-a)], so 

               f(b)-f(a)
g'(c) = f'(c)- ----------  = 0
                 b-a

Therefore, 

          f(b)-f(a)
f'(c) =  ----------
            b-a


That proves the theorem!

You can find the proof here, the presentation might be better on those pages:

The Proof of the Mean Value Theorem
http://www.millersv.edu/~bikenaga/calculus/mvt/mvtpf.html

and here:

The Mean Value Theorem
http://www.math.hmc.edu/calculus/tutorials/mean_value/


I hope this helps. If you require any clarifications please do not hesitate to ask.

Thanks.
endo


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