**Problem** (T. M. Flett, 1958). Let be a differentiable function. Show that if then there exists such that

**Solution**. We consider two cases.

Case 1 : Define the function as follows

Clearly is continuous on and differentiable on Also for we have

So we are done if we show that for some To do that, we consider the three possibilities

If then and thus, by Rolle’s theorem, there exists such that

Now suppose that Then, by

and so we cannot have that for all because then we would have the contradiction Thus there exists such that Now consider the function which is clearly continuous because is continuous. We have and Thus, by the intermediate theorem, there exists such that So and hence, by Rolle’s theorem, there exists such that as desired. The proof for the case is similar.

Case 2 : The general case. Consider the function Then and so Thus, by Case 1, there exists such that and so

which simplifies to

That in Flett’s mean value theorem is not necessarily unique. In fact, there could be many of them, as the following example shows.

**Example 1**. Consider the function For any integer let be the number of solutions of the equation

in the interval Show that

**Solution**. We have Let We want to find the number of solutions of in the interval We have

Since and on is increasing on and so it has no root in that interval.

Now we count the number of roots of in those subintervals of which are in the form for some integer First see that if is even, then and if is odd, then So, by the intermediate value theorem, has a root in Since for all cannot have more than one root in So has exactly one root in

Finally, we look for possible roots of in the subinterval Well, if is odd, then and So in this case, has a root. This root is unique because has no root in But if is even, then and on Thus on and thus has no root in this case.

So we have shown that has no root in but it has exactly one root in each subinterval That gives roots of

We have also shown that, in has no root if is even and has exactly one root if is odd. So the number of roots of in is for even and is for odd.

**Example 2**. Let be a continuous function and suppose that Show that there exist such that

**Solution**. Consider the function

See that and Thus, by Rolle’s theorem, there exists such that and so

Now, since Flett’s mean value theorem gives a that satisfies the condition i.e.

and hence

Now let See that and Thus, by Rolle’s theorem, there exists such that and so

Similarly, let Then and so for some implying that

**Execsrie 1**. Geometrically, what does Flett’s mean value theorem mean?

**Exercise 2**. Let be a differentiable function. Show that if then there exists such that

**Exercise 3**. Think about this: is there a differentiable function that satisfies the conditions i) and ii) there are infinitely many for which If there is, give one!