Consider the following statement of the intermediate value theorem for derivatives:
Assume is differentiable on an open interval . Let be two points in . Then, the derivative takes every value between and somewhere in .
- Define a function
Prove that takes every value between and in the interval . Then, use the mean-value theorem for derivatives to show takes all values between and somewhere in the interval .
- Define a function
Show that the derivative takes on all values between and in the interval . Conclude that the statement of the intermediate-value theorem is true.
- Proof. First, since is differentiable everywhere on the interval , we know is continuous on and differentiable on . Thus, if and then is continuous at since it is the quotient of continuous functions and the denominator is nonzero. If then
hence, is continuous at as well. Therefore, is continuous on the closed interval . So, by the intermediate value theorem for continuous functions we know takes on every value between and somewhere on the interval . Since this means takes on every value between and somewhere on the interval .
By the mean-value theorem for derivatives, we then know there exists some such that
for some . Since , we then conclude there is some such that for any . Since takes on every value between and , so does .
- Proof. This is very similar to part (a). By the same argument we have the function is continuous on ; thus, takes on every value between and by the intermediate value theorem for continuous functions. Then, by the mean value theorem, we know there exists a such that
Thus, takes on every value between and . Since ; takes on every value between and
Let be an integrable, non-negative function on the interval . Prove that if
then at each point at which is continuous.
Proof. The proof is by contradiction. Let be a point at which is continuous, and suppose . This implies since is non-negative by hypothesis. Since is continuous at , we know by the sign-preserving property of continuous functions (Theorem 3.7 in Apostol), that there is some neighborhood about , say such that has the same sign as for all , i.e., such that for all . But then by the monotone property of the integral we know,
This is a contradiction since is nonnegative we know
Since is strictly positive, the sum of the three pieces of the integral cannot equal
Define functions and as follows:
Find a formula for . Find the values at which is continuous.
If , we have and we compute the composition,
If , then and we compute the composition,
This is the same function as in this previous exercise, and it is continuous everywhere.
Given a function and a closed interval such that
for all .
- Prove is continuous at every point .
- If is integrable on prove
- If is any point prove
- Proof. Let be any point in . Then, for any let . Then
Thus, for any , we have whenever . Hence,
Therefore, is continuous at every point
- Proof. Since is a constant (the value of the function evaluated at the constant ), we have
Therefore, we can compute,
The final inequality follows since, using the monotone property of the integral,
Then, since by our hypothesis on , we have
- Proof. The proof proceeds similarly to that of part (b),
Then, since , we break the integral into two pieces (to deal with the fact that for and for ):
But now, we know implies , so we have the inequality,