# Mean value theorem, closed intervals

1. Jun 26, 2015

Mean Value Theorem

Suppose that $f$ is a function that is continuous on $[a,b]$ and differentiable on $(a,b)$. Then there is at least one $c$ in $(a,b)$ such that:
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

My question is: wouldn't it be better to state that $c$ is in $[a,b]$ rather than $(a,b)$? For example, if $f(x) = 2$ for $1 \leq x \leq 3$, then:
$$f'(x) = 0 = \frac{f(3) - f(1)}{2}$$
For all $x$, including 3, which is one of the endpoints of the interval.

2. Jun 26, 2015

### micromass

Staff Emeritus
If $c$ is in $(a,b)$, then $c$ is in $[a,b]$. Besides, $f$ might not be differentiable at $a$ or $b$.

3. Jun 26, 2015

### verty

Because the theorem is a stronger result.

4. Jun 26, 2015

Yeah, I think that's it. $f$ is differentiable on $(a,b)$, not $[a,b]$.