4.1 Extreme Values of a Function by SGLee - HSKim-SWSun
1- 4. Determine if the following statement is true or false. Explain your answer.
1. If is a continuous function, then has a maximum at only one point in .
False. Consider that has more than one maximum.
2. One can apply the Mean Value Theorem to on .
False, since is not differentiable on the open interval .
3. If a continuous function has a extreme value on , then has a absolute maximum or minimum value on .
False. Draw the graph of on .
4. For a continuous function , has only one zero provided is strictly decreasing.
True. We may prove it By the Way Of Contradiction (BWOC).
5- 8. Find all critical numbers of given functions.
Therefore critical numbers of are .
To find critical numbers, we differentiate .
Therefore, and (since ).
So, critical numbers of are or .
9-13. Find all local maxima and minima if there exist.
No local minimum and maximum when
Local minimum , but no local maximum.
x=0 is a critical point, but have not local maximum or local minimum.
Thus, when x=pi/2, f(pi/2)=1 is a local maximum.
if , , then has a local minimum at .
if , , then has a local minimum at .
if , , then has a local maximum at .
then has a local maximum at
14-17. Find the intervals where the function is increasing or decreasing.
Decreasing on , increasing on
Increasing on , decreasing on
Answer : Increasing on , decreasing on
17. Suppose and for all . Using the Mean Value Theorem show for all .
By the Mean Value Theorem,
Since , we have
18-20. Prove the inequality.
Let , , . Then , ,
From the inequality and the Mean Value Theorem,
, . Therefore for all .
So, for all .
21. Let . Find the number of zeros and give a proper interval containing all zeros.
7 zeros in
22. Show that for .
23-24. Prove the inequality using the Mean Value Theorem.
Let , then by the Mean Value Theorem there exists in such that
. Since for , we have
Let , than there exists in such that
, we have .
25. Prove that the inequality for all and using the Mean Value Theorem.
Let , , than there exists in such that
26. Let be a continuous function. Prove that the only functions satisfying for all are of the form , by the Mean Value Theorem.
Applying Mean Value Theorem to on , there exists in such that
for all . This implies
Now set , , then for all .
27. Suppose is continuous on and differentiable on . Prove that there exists such that .
Let , then
By Rolle's theorem, here exist such that , so
28. Prove Fermat's Theorem: If is differentiable at and has an extreme value at , then .
Suppose, for the sake of definiteness, that has a local maximum at . Then, if is sufficiently close to .
This implies that if is sufficiently close to 0, with being positive or negative, then and therefore .
We can divide both sides of this inequality by a positive number. Thus, if and is sufficiently small, we have
Taking the right-hand limit of both sides of this inequality, we get
Since exists, we have
and so we have shown that .
If , then the direction of the inequality (1) is reversed. We have
So, taking the left-hand limit, we have
We have shown that and also that . Since both of these inequalities must be true, the only possibility is that .
29. Prove Rolle's Theorem. (http://en.wikipedia.org/wiki/Rolle's_theorem)
Since , we can use the above result due to Farmat’s Theorem to conclude this.
Suppose that the maximum is obtained at an interior point of . We shall examine the above right- and left-hand limits separately.
For a real such that is in , the value is smaller or equal to because attains its maximum at . Therefore, for every , , hence
, where the limit exists by assumption (it may be minus infinity).
Similarly, for every , the inequality is reversed because the denominator is now negative and we get
, where the limit might be plus infinity.
Finally, because the above right- and left-hand limits agree, the derivative of at must be zero.
30. Prove the Increasing and Decreasing Test.
(a) Let and be any two numbers in the interval with . According to the definition of an increasing function we have to show that .
Applying the Mean Value theorem to on the interval , there exists such that .
By assumptions, and , we get . In particular, . Since and were arbitrary, we conclude that whenever .
31. Prove the cases (ii), (iii) of the First Derivative Test. (http://en.wikipedia.org/wiki/First_derivative_test)
(ii) Let us choose sufficiently near to such that . By the Mean Value Theorem, there exist with and such that
This implies for and for . In particular is decreasing on and increasing on . Hence has a local minimum at .
(iii) Let us choose sufficiently near such that . By the Mean Value Theorem, there exist with and such that
Since for and for . Hence has neither local maximum nor local minimum at .
32. Prove the case (b) of the Second Derivative Test. (http://en.wikipedia.org/wiki/Second_derivative_test)
(b) Suppose we have . Then
Thus, for sufficiently small we get
which means that if so that is decreasing to the left of , and that if so that is increasing to the right of .
Now, by the first derivative test we know that has a local maximum at .
First ICM, Zurich, 1932