Calculus-Sec-4-1-Solution


4.1    Extreme Values of a Function          by SGLee - HSKim-SWSun

                                                                                                            http://youtu.be/_V4MryNEzWY

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. .

http://matrix.skku.ac.kr/cal-lab/cal-4-1-5.html 

 




6. 

 

 

 

Therefore critical numbers of  are .




 7. .

http://matrix.skku.ac.kr/cal-lab/cal-4-1-7.html 

 

 




 8. .

http://matrix.skku.ac.kr/cal-lab/cal-4-1-8.html 

 

 




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. .

http://matrix.skku.ac.kr/cal-lab/cal-4-1-9.html 

 

 No local minimum and maximum when 




 10. .

http://matrix.skku.ac.kr/cal-lab/cal-4-1-10.html 

 

 Local minimum , but no local maximum.




 11. .

http://matrix.skku.ac.kr/cal-lab/cal-4-1-11.html 

 













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. 

http://matrix.skku.ac.kr/cal-lab/cal-4-1-15.html 

 

 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.

http://matrix.skku.ac.kr/cal-lab/cal-4-1-20.html 

 

 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 .

     http://en.wikipedia.org/wiki/Fermat'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. (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 

 

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