Prove that for all we have
Before proceeding with the proof, we recall the second mean-value theorem for integrals (Theorem 5.5 on p. 219 of Apostol). For a continuous function on the interval if has a continuous derivative which never changes sign on the interval then there exists a such that
Proof. Now, we want to apply the mean-value theorem above with and . Since is continuous everywhere , it is continuous on any interval . Then,
is continuous for all (so, in particular, for all ). Furthermore, since for all we have that for all . Thus, is continuous and never changes sign in any interval . Therefore, we can apply the mean-value theorem to conclude there exists a for any such that
But since for all we know for any and . Hence,
In each of the following cases find a function with continuous second derivative satisfying the given conditions.
- for all , , and .
- for all , , and .
- for all , , and for all .
- for all , , and for all .
- There can be no function meeting all of these conditions since implies is an increasing function (since its derivative, , is positive). But then contradicts that is increasing.
- Let . Then
Furthermore, for all .
- There can be no function meeting all of these conditions. Again, for all implies that is increasing for all . Therefore, implies for all . Then, by the mean-value theorem, we know that for any there exists some such that
Now, choose . Then, , contradicting that for all .
- We’ll define piecewise as follows
Then, we can take the derivative of each piece (and see that they are equal, so the derivative is defined at )
Taking the derivative again we find
Thus, for all and . Furthermore, for we have
Prove the following inequalities using the mean value theorem:
- , for .
- Proof. Define and . Then and are continuous and differentiable everywhere so we may apply the mean value theorem. We obtain
The final step follows since for all
- Proof. Let , , then and . So, by the mean-value theorem we have there exists a such that,
But, since is an increasing function on the positive real axis, and we have we know
Further, since and is positive we can multiply all of the terms in the equality by without reversing inequalities to obtain,
Therefore, substituting from above we obtain the requested inequality:
Consider a polynomial . We say a number is a zero of multiplicity if
- Prove that if the polynomial has zeros in , then its derivative has at least zeros in . More generally, prove that the th derivative, has at least zeros in the interval.
- Assume the th derivative has exactly zeros in the interval . What can we say about the number of zeros of in the interval?
- Proof. Let denote the distinct zeros of in and their multiplicities, respectively. Thus, the total number of zeros is given by,
By the definition given in the problem, if is a zero of of multiplicity then
Taking the derivative (using the product rule), we have
Thus, again using the definition given in the problem, is a zero of of multiplicity .
Next, we know from the mean-value theorem for derivatives, that for distinct zeros and of there exists a number (assuming, without loss of generality, that ) such that . Hence, if has distinct zeros, then the mean value theorem guarantees numbers such that . Thus, has at least:
By induction then, the th derivative has at least zeros.
- If the th derivative has exactly zeros in , then we can conclude that has at most zeros in .
Prove that the expression
is an equivalent form of the mean-value theorem.
Find the value of in terms of and when:
For parts (a) and (b) keep fixed with and find the limit of as tends to 0.
Proof. If is continuous on and differentiable on , then by the mean-value theorem we have
Letting and for some (since ), we have
(This follows since from our definitions, is the distance from . Then, since is somewhere in the interval its value must be plus some portion of the distance to . This portion is then , which is how we know .) Substituting and and ,
Now for parts (a) and (b).
- If , we have , so,
- If , we have . So,
Consider the equation
Show that there are two values of such that the equation is satisfied.
Proof. Let . (We want then to find the zeros of this function since these will be the points that .) Then,
Since for any (since ), we have
Then, is continuous and differentiable everywhere, so we may apply Rolle’s theorem on any interval. So, by Rolle’s theorem we know has at most two zeros (if there were three or more, say and , then there must be distinct numbers and with such that , but we know there is only one such that ).
Furthermore, has at least two zeros since , , and . Thus, by Bolzano’s theorem there are zeros between each of these points. We have that the number of zeros of is at most two and at least 2. Hence, the number of zeros must be exactly two