The mean value theorem is an essential result in calculus and real analysis that expresses a relationship between the derivative of a function and its average rate of change. More precisely, the mean value theorem states that if a function is continuous over a closed interval \(I = [a,b]\) and differentiable on its interior \((a,b)\), then there is a point in its interior such that the function’s derivative at that point is equal to the average rate of change of the function over \(I\). In this post, we prove a generalization of the mean value theorem and discuss applications of the mean value theorem.

#### Generalized mean value theorem

**Result** (generalized mean value theorem). If \(f\) and \(g\) are continuous on \([a,b]\) and differentiable in \((a,b)\), then there is a point \(c \in (a,b)\) such that $$(f(b) – f(a)) g'(c) = (g(b) – g(a)) f'(c).$$

The proof is as follows:

Define \(h: [a,b] \rightarrow \mathbb{R}\) by $$h(x) = (f(b) – f(a)) g(x) – (g(b) – g(a)) f(x).$$ It is clear that \(h\) is continuous on \([a,b]\) and differentiable in \((a,b)\) with \(h(a) = h(b)\). Therefore, by Rolle’s theorem, there is a point \(c \in (a,b)\) such that \(h'(c) = 0\). Hence, for some \(c \in (a,b)\) $$(f(b) – f(a)) g'(c) = (g(b) – g(a)) f'(c).$$

This completes the proof of the generalized mean value theorem.

**Remark**. The generalized mean value theorem has many applications and can be used to prove L’Hôpital’s rule. In the next section, we explain how the mean value theorem which is a special case of the generalized mean value theorem has applications in integral calculus.

#### The mean value theorem

If in the generalized mean value theorem, we put \(g(x) = x\), then we obtain the mean value theorem:

**Result** (the mean value theorem). Let \(f\) be continuous on \([a,b]\) and differentiable in \((a,b)\). Then, there is a point \(c \in (a,b)\) such that $$f'(c) = \frac{f(b) – f(a)}{b-a}.$$ In other words, if a function is continuous over a closed interval \(I = [a,b]\) and differentiable on its interior \((a,b)\), then there is a point in its interior such that the function’s derivative at that point is equal to the average rate of change of the function over \(I\).

#### Some applications of the mean value theorem

In the following, with the help of the mean value theorem, we prove two results that are essential in the integral calculus:

**Result** (a characterization of constant functions). If the derivative of a function vanishes over an interval \(I\), then it is a constant function over that interval \(I\).

The proof goes as follows:

Let \(a < b\) be arbitrary real numbers in the interval \(I\) . By the mean value theorem, there is a \(c\) in the open interval \((a,b)\) such that $$f'(c) = \frac{f(b) – f(a)}{b-a}.$$ Since \(f'(c) = 0\), we have \(f(b) = f(a)\). Hence, \(f\) is a constant function and the proof is complete.

**Result**. Let \(f\) and \(g\) be differentiable over an interval such that their derivatives are equal over that interval. Then, their difference is a constant real number.

**Proof**. Set \(h = f-g\) and apply the above result for \(h\).

**Result** (a criteria for monotonicity). Let \(f\) be differentiable in the open interval \((a,b)\). Then, the following statements hold:

- If \(f’\) is positive, then \(f\) is strictly increasing, i.e. \(x_2 > x_1\) implies that \(f(x_2) > f(x_1)\), for all real numbers \(x_1\) and \(x_2\).
- If \(f’\) is negative, then \(f\) is strictly decreasing, i.e. \(x_2 > x_1\) implies that \(f(x_2) < f(x_1)\), for all real numbers \(x_1\) and \(x_2\).
- If \(f’\) is non-negative, then \(f\) is increasing, i.e. \(x_2 \geq x_1\) implies that \(f(x_2) \geq f(x_1)\), for all real numbers \(x_1\) and \(x_2\).
- If \(f’\) is non-positive, then \(f\) is decreasing, i.e. \(x_2 \geq x_1\) implies that \(f(x_2) \leq f(x_1)\), for all real numbers \(x_1\) and \(x_2\).

Proof. Let \(x_1\) and \(x_2\) be two distinct real numbers. Then, by the mean value theorem, there is a point \(c\) between them such that $$f'(c) = \frac{f(x_2) – f(x_1)}{x_2-x_1}.$$

In the post on uniform continuity, we use the mean value theorem to prove that if the derivative of a function is bounded, then it is uniformly continuous.