Calculus-Sec-4-1-Solution

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.

5. .

6.

Therefore critical numbers of  are .

To find critical numbers, we differentiate .

Then  .

Therefore,  and  (since ).

So, critical numbers of  are  or .

9-13. Find all local maxima and minima if there exist.

9. .

No local minimum and maximum when

10. .

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.

12.

if , then  has a local minimum at .

if , then  has a local minimum at .

if , then  has a local maximum at .

13.

then  has a local maximum at

14-17. Find the intervals where the function is increasing or decreasing.

14.

Decreasing on ,  increasing on

15.

Increasing on , decreasing on

16.

Answer :  Increasing on , decreasing on

17. Suppose  and  for all . Using the Mean Value Theorem show  for all .

By the Mean Value Theorem,

and

Since  , we have

.

18-20. Prove the inequality.

18.

So,  for

19.  ,

Let . Then ,

From the inequality  and the Mean Value Theorem,

. Therefore  for all .

20.

and

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 .

Then, .

23-24. Prove the inequality using the Mean Value Theorem.

23. .

Let , then by the Mean Value Theorem there exists  in  such that

. Since   for , we have

.

24.

Let , than there exists  in  such that

. Since,

, we have .

25. Prove that the inequality  for all  and  using the Mean Value Theorem.

Let , than there exists  in  such that

Thus,.

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

hence

,  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

and .

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

and .

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