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. . 






Therefore critical numbers of  are .

 7. . 



 8. . 



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.

 11. . 


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.







      So,  for 

19.  , 



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 .






      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




 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





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 .'s_theorem_(stationary_points)



      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. ('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. (



      (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. (


      (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 


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