- Let be a function with second derivative continuous and nonzero on an interval . Furthermore, let be a constant such that
Use the second mean-value theorem for integrals (Theorem 5.5 in Apostol) to prove the inequality
- If prove that
- Proof. Since we have the assumption that for all we may divide by , to obtain
Then, to apply the second mean-value theorem for integrals (Theorem 5.5 of Apostol) we define functions
The function is continuous since is a composition of continuous functions (we know is continuous since it is differentiable) and is continuous (again, it is differentiable since exists and is continuous by assumption). Then the product of continuous function is also continuous, which establishes that is continuous. We also know that meets the conditions of the theorem since it has derivative given by
This derivative is continuous since and are continuous and is nonzero. Furthermore, this derivative does not change since on since is nonzero on (and by Bolzano’s theorem we know that a continuous function that changes sign must have a zero). Therefore, we can apply the second mean-value theorem:
- Proof. Using part (a), we take , giving us
Thus, is continuous and never changes sign. Furthermore, (where is a given constant) and since . Thus,
Hi, this resource was a great find!
One thing left me stumped working on this. On the second row of the solution to part one, I initially summed the two integrals, getting 2/m rather than 4/m.
I figured that I could use the triangle inequality; but I can’t understand if summing the two integrals gives an incorrect result. Could you explain?
On second thoughts… you can’t actually sum the two integrals when 1/(phi’) is actually evaluated at a and b. But the two integrals can still both be equal to 2.