Consider the function on 0123 defined by gx 1 0 x 1 1 2

Consider the function on (0,1)(2,3) defined by

g(x) ={ 1, 0 < x < 1

{1, 2 < x < 3 .

Then g\'(x)=0 on its domain but g is not a constant function. Why does this not contradict the mean value theorem?

Solution

1. The mean value theorem (MVT) requires continuity of the function on its domain.

2. Moreover the MVT requires the function to be defined and continuous on a closed interval.

3. Here the given function is not continuous on its disconnected domain achieving two different constant values on each of its domain\'s constituent parts. Rather it\'s piecewise continuous. Also the domain of the function is a union of two open intervals.

4. Thus the function has first derivative 0 for all x in its domain without being a constant function but doesn\'t satisfy the conditions of the MVT.

5. It\'s important to understand that closeness of the interval that plays as the domain of the function along with the continuity of the function is a necessary condition so that the function can achieve its extremum on the domain and also to become defined on the end points.

That is why this example doesn\'t contradict the Mean Value Theorem.