Calculus-Sec-2-2-Solution


  

  2.2   Continuity                                        by SGLee - HSKim -SWSun

                                                                   http://youtu.be/azrkT1RP4-chttp://youtu.be/hj8d-j_DGf4 

 

 1. Given continuous functions  and  with  and , find .

http://matrix.skku.ac.kr/cal-lab/cal-2-2-1.html 

 

    Since  and  are continuous functions.

       

         




2. Given continuous functions  and  with  and , find .

   

   Since  is continuous, .




 3. Show that the function is discontinuous at .

http://matrix.skku.ac.kr/cal-lab/cal-2-2-3.html 

 

      We may draw  with the following Sage command. It shows the function diverges

      (+infinity) at . This shows f(x) is discontinuous at .

 If we use the following Sage command, we may get limit(abs(ln((x-2)^2))) at  directly.

 It shows  is discontinuous at .




4-7. Determine the points of discontinuity of . At which of these numbers is  continuous from the right, from the left or neither? Sketch the graph of .

 

 4. 

http://matrix.skku.ac.kr/cal-lab/cal-2-2-4.html 

 

      We may draw  with the following Sage command. It shows the function is discontinuous at .




5. 

 

We see that  exists for all  except . Notice that the right and left limits are different  .                                                             




6. 

 

    We see that  exists for all  except  and . Notice that the right and left limits are different at  and .                                           




7. 

     We see that  exists for all  except  and . Notice that the right and left limits are different at  and .




8-10. For what values of the constant  is the function  continuous on ?

 

8. 

 

    

   Thus, for  to be continuous on 




 9. 

http://matrix.skku.ac.kr/cal-lab/cal-2-2-9.html 




10. 

http://matrix.skku.ac.kr/cal-lab/cal-2-2-10.html 

  

        Suppose that function  is continuous on  . Find C such that 

       Find C such that   is continuous at  (But  is not continuos at ). The solution is .




11-13. Show that the following functions  has removable discontinuity at . Also find a function  that agrees with  for  and is continuous on ℝ.

 

11. 

 http://matrix.skku.ac.kr/cal-lab/cal-2-2-11.html

     does not exist. Hence  has removable discontinuity at . Then we can change the expression to remove the discontinuity.

     Let . It agrees with  for  and is continuous on ℝ.




12. 

      for .

    The discontinuity is removable and  agrees with  for  and is continuous on ℝ.




13. 

      for .

    The discontinuity is removable and  agrees with  for  and is continuous on ℝ.




14. Let . Is  removable discontinuity?

 

     Since ,

       is not a removable discontinuity.




15. If , show that there is a number  such that .

 

     is continuous on the interval  and .

     Since , there is a number  in  such that  by the Intermediate Value Theorem.




16. Prove using Intermediate Value Theorem that there is a positive number  such that .

     Let . We know that  is continuous on the interval  and . Since ,

     there is a number  in  such that  by the Intermediate Value Theorem.




17-22. Prove that there is a root of the given equation in the specified interval by using the Intermediate Value Theorem.

 

 17. 

http://matrix.skku.ac.kr/cal-lab/cal-2-2-17.html 

 

        Let . Then .

        Since  is a real-valued continuous function on the interval , there is a root  such that  by the intermediate value theorem.







 18. 

http://matrix.skku.ac.kr/cal-lab/cal-2-2-18.html  

         Let . Then .

        Since  is a real-valued continuous function on the interval , there is a root  such that  by the intermediate value theorem.










 19. 

http://matrix.skku.ac.kr/cal-lab/cal-2-2-19.html 

 

        Let . Then .

        Since  is a real-valued continuous function on the interval , there is a root  such that  by the intermediate value theorem.







 20. 

http://matrix.skku.ac.kr/cal-lab/cal-2-2-20.html 







21. 

 

       is continuous on the interval  and .

      Since , there is a number  in  such that  by the Intermediate Value Theorem.

      Thus, there is root of the equation  in the interval .




22. 

 

      is continuous on the interval  and .

     Since –6.696<0<e, there is a number  in  such that  by the Intermediate Value Theorem. Thus, there is root of the equation   in the interval .




23-26. Show that each of the following equations has at least one real root.

 

 23. 

http://matrix.skku.ac.kr/cal-lab/cal-2-2-23.html  

 http://math2.skku.ac.kr/home/pub/37

      We may draw both  and  in one graph to find intersections. It shows the function has two real roots in   and .




The following Sage commands give the value of the intersection in each interval,  or .







24. 

     Let . Then   and . So by the Intermediate Value Theorem, there is a number  in (-1,0) such that .

     This implies that .




25. 

     Let . Then  and . So by the Intermediate Value Theorem, there is a number  in (1,2) such that .

     This implies that 




26. 

     Let . Then  and , and  is continuous in 

    So by the Intermediate Value Theorem, there is a number  in  such that . This implies that .




27-28. Find the values of  for which  is continuous.

 

27. 

 

     BWOC. Suppose  is continuous at  in ℝ, then .

       But if  is rational, then  but  or  can be 1. if  is irrational, then  but  or  can be 0. It occurs a contradiction. So such value  in ℝ exist.




28. 

 

     is continuous only at  since . But there is no nonzero  in ℝ. such that .




                                    

 

                                            

                                                                            2002 FIFA World Cup

 

                                                                       Back to Part I