Continous function on [a,b] not differentiable at only one point in (a,b) not satisfying Lagrange Mean value theorem

Question

Give an example of a continuous function on $[a,b]$ not differentiable at only one point in $(a,b)$ not satisfying Lagrange Mean value theorem.

Answer 1

For $f(x)=|x|$ we have $$\frac{f(1)-f(-1)}{1-(-1)}=0$$ while $f'(x)\in\{-1,1\}$ for any $x\ne 0.$

Answer 2

The first thing what you should do is to visualize what the theorem states.

In short, the theorem says that there is at least one point in the interval $(a,b)$ such that the derivative at that point is the same as the derivative of the linear segment connecting the points $(a,f(a))$ and $(b,f(b))$.

In a practical point of view, the theorem works because it is assumed that the function is 'smooth', just try to find a smooth function by making some sketches. Please do this first before you continue reading my answer. This will also help you by understanding the theorem better. Of course, you will see that you will not be able to construct a smooth function that does not satisfy the theorem but that does not really matter.

Let's say you constructed an example with $a=-1$, $b=1$ and $f(a)=f(b)=0$. You could have connected the two points by a straight line, or maybe a polynomial of a higher order. In both cases, you will always end up with a point $c$ in between $a$ and $b$ such that $f'(c)=0$, exactly what you don't want to have. Now, what you should do is to make a function which does not have a derivative of $0$ in between the points. Or, as you could rephrase, find a function such that it attains a maximum between $a$ and $b$ (just one local maximum) but such that it is not differentiable at that point.

It should now be obvious that the triangular function $\Lambda(t)$ works but please note that you could find a whole lot of other different functions here.