# mean value theorem differentiability hypothesis violation

The Mean Value Theorem states that if a function $f$ is continuous on the closed interval $[a, b]$, where $a < b$, and differentiable on the open interval $(a, b)$, then there exists a point $c$ in $(a, b)$ such that $$f'(c) = \frac{f(b)-f(a)}{b-a}.$$

Apparently this theorem holds true only when the function in question is both differentiable and continuous, I tried to think about it as a car travelling at a certain average speed would mean that it certainly touches its mean value at least once. I am not able to understand why this is not true if a function is continuous but not differentiable, shouldnt continuity be the only factor? Why does i need it to be differentiable?

• what on earth would $f'$ be if the function is not differentiable??? – Ittay Weiss Oct 19 '15 at 9:27
• if not differentiable then it wont hold for min or max. remember min and max comes from this mvt. – d13 Oct 19 '15 at 9:28
• Here's an example where the function is continuous on the whole closed interval and differentiable at all but one point in the open interval, yet there is no $c$ as claimed in the MVT. That is, even a very slight failure of differentiability ruins the theorem. Let $a=-1$, $b=1$, and $f(x)=|x|$. – Andreas Blass Oct 19 '15 at 9:42
• Allright thanks , but would this mean that the function doesnt even touch the mean value or does it just mean that at the mean value the slope is not f(b)-f(a)/b-a. – Tac Oct 19 '15 at 12:35

If function is not differentiable, then $f'(c)$ is meaningless.
• Mean value theorem. Slope of the blue line is $(f(b)-f(a))/(b-a)$. Slope of the green line is $f'(c)$ and it is equal to slope of blue line. It's easy to see that when a function is differentiable we can always find such a green line. – luka5z Oct 19 '15 at 9:35