1. ## Clarification needed

I've had no issues employing the following theorem but there's just a slight bump that need's ironing out with the hypotheses and the domain.

Mean Value Theorem

If f(x) is continuous in the interval $\displaystyle a \leq x \leq b$ and if f'(x) exists at each value of x for which $\displaystyle a < x < b$, then there exists at least one value c of x between a and b such that $\displaystyle f(b) - f(a) = f'(c)(b - a)$

Why are the endpoints omitted for the interval of the derivative? The only thing I can come up with is that c is between a and b. So f can be continuous at a and b but not necessarily differentiable at a and b? But if f happened to be a straight line it would have to be differentiable at a and b.

Since continuity is contained within differentiability, then would it not be more concise to say

If f(x) is differentiable on [a,b], then there exists at least one value c of x between a and b such that $\displaystyle f(b) - f(a) = f'(c)(b - a)$

2. ## Re: Clarification needed

The version of the "mean value theorem" you give "If f(x) is differentiable on [a,b], then there exists at least one value c of x between a and b such that f(b)−f(a)= f′(c)(b−a)" would be true but not sufficiently general. if a function is continuous on [a, b] and differentiable on (a, b) but not at a or b the conclusion is still true. For example, If f(x)= x- x^2 for x greater than or equal to 0 and less than or equal to 1, f(x)= -x for x<0, f(x)= x- 1 for x> 1. f is continuous for all x, and in particular on the closed interval [0, 1]. It is differentiable on the open interval (0, 1) but not at x= 0 or x= 1. The 'mean value' between 0 and 1 is (f(1)- f(0))/(1- 0)= 0 and the derivative at x= 1/2 is - 2(1/2)= 0.

Continuity at the endpoints is important because the mean value theorem specifically uses those values. Consider the same example as above, but with f(0)= 1. Now (f(1)- f(0))/(1- 0)= -1 but there is no x in (0, 1) where the derivative, 1- 2x, is equal to -1.

3. ## Re: Clarification needed

I've drawn a diagram of the function for x<0, 0<= x <=1, x > 0, and yes I can see the derivative is not unique at the endpoints. My statement above was too general I see that now thanks. In your last comments, the mean value theorem failed because f(0) = 1 was not an endpoint of the function it was just redefined for an arbitrary value.

Just one last thing, the intervals for the function you gave presumably continuous passing through zero and out through 1, shouldn't we put f(x) = -x, x<=0 and f(x) x -1, x =>1 just so it's defined at those values when we descrive f(x) differently?