Let . Iff is continuous on and differentiable on , then the mean value theorem for derivatives states that  , there is such that
If f is continuous on , the mean value theorem for definite integrals states that there is such that
Generally speaking, the c in the mean value theorem for the derivative need not equal the in the mean value theorem for the definite integral. It is possible that , for example, . In this case, . In this note, we determine all functions for which the two mean value theorems give the same c.
Our results rely heavily on the following identity theorem for real analytic functions, its proof can be found in .
Theorem If f and g are real analytic functions on an open interval U and there is an open set such that
2. Main Result
Lemma 1 Suppose f is analytic on an open interval U, f and have inverses and for any with , there is such that
then , where , and are arbitrary real numbers.
Proof. By the hypothesis,
Fix . If we denote , then and the above relation becomes
Taking the derivative of both sides, we obtain
To compute , we use the derivative of an inverse function and the chain rule,
Replacing by gives,
We need to find f satisfying the above relation. Let W be an open neighborhood of a such that
Then (1) becomes
Combining like terms and cross multiplying, we obtain
If , then it is easy to see that . In this case, , which cannot happen because we assume that f is invertible. Thus, . Equating the respective coefficients of and on both sides of (2), we obtain
In general, for , , for some . To find , we observe that satisfies (1). Hence the power series of should satisfy (2). But in the power series expansion of at , the coefficient .
That is, . In other words, for f to satisfy (1), the power series expansion of at must be
where are real numbers. If , then , violating the fact that has an inverse. Thus,
By the identity theorem, on U.
Finally, we can relax the requirement that both f and have inverse.
Theorem 1 If f is analytic on an open interval U, and with , then the two mean value theorems give the same c if and only if or , where , and are any real numbers.
Proof. The necessary condition is obvious. We only prove the sufficient condition. If f is constant on U, then we are done. Suppose f is not constant on U, then there is some such that . Since is continuous at , there is an open interval I of with such that on I. Hence, exists on I.
Case 1. is constant on I, then on I, for some with . By the identity theorem, on U.
Case 2 is not constant on I, i.e. for some . But is continuous at , so there is an open interval J containing with such that on J. Then, bothf and have inverses on J, and by the Lemma above, on J, for some , and . But the identity theorem implies that, on U.
In this work, we find that if and f is an analytic function, there is satisfies both
then f must be a linear function or an exponential function. The proof relies heavily on the fact that f is analytic, we don’t know if this condition can be weakened.
The author is indebted to Alexander Kasiukov for many helpful discussions, without which this project would not be possible