1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Intermediate Value Theorem

  1. Mar 2, 2012 #1
    1. The problem statement, all variables and given/known data
    I attached my problem as a JPEG file. The black ink is my own work, and the green ink is my teachers note's. If you are confused by the part that states these two steps prove continuity, I considered the evaluation of the function at a value and the evaluation of the limit at the same value to be one step, that's why there is two.
    Well, I asked my teacher to write the proper way to write the solution to this sort of problem, but evidently she did not. So, I was wondering if someone could possibly interpret her notes, and also if someone could show me the mathematically proper way to solve this problem.



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 2, 2012 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You only showed that the function is continuous at -3 and +3, not on the whole interval [-3,3].

    Don't you have any theorems you can cite, to the effect that every polynomial is continuous at all points?
     
  4. Mar 2, 2012 #3
    Doesn't taking the limit of -3 from the right side and the limit of 3 from the left side prove continuity of the interval?
     
  5. Mar 2, 2012 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, it just proves continuity (from the right) at -3, and continuity (from the left) at +3.

    If you were to change the value of your function at say, x = 0, so that a discontinuity was formed, your calculations would not capture this fact.
     
  6. Mar 2, 2012 #5

    Mark44

    Staff: Mentor

    No, not at all. In fact, this doesn't prove much of anything.

    You're confusing the concepts of the limit of a function at a point and the continuity of a function.

    For a function f, if the two one-sided limits exist at a number a and are equal, then the two-sided limit exists. IOW, if
    [tex]\lim_{x \to a^-} f(x) = L, \text{ and }\lim_{x \to a^+} f(x) = L[/tex]
    then
    [tex]\lim_{x \to a} f(x) = L[/tex]

    A function f is continuous at a number a in its domain provided that
    [tex]\lim_{x \to a} f(x) = f(a)[/tex]
     
  7. Mar 2, 2012 #6
    Well, the limit is taking values that become increasingly closer to, for instance, -3; so if we are taking values from the right of -3 that would be all the values of (-3, 3], but since it's a polynomial I can just simply input the value -3 three. I don't see how it doesn't prove continuity. That is what my professor taught; to find continuity, she said you need to find out if the limit and the function equal each other.
     
  8. Mar 2, 2012 #7

    Mark44

    Staff: Mentor

    No, it wouldn't. The limit involves values of x that are close to -3, and just to the right of it.
    That's continuity at a point, which I mentioned in my last post. For a function to be continuous on an entire interval, it has to be continuous at each point in the interval, and you didn't show that.
     
  9. Mar 2, 2012 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, what your professor said is true- to prove that a function is continuous at x= a show that [itex]\lim_{x\to a} f(x)= f(a)[/itex]. What you have done is show that this function is continuous at x= -3 and at x= 3, not at every point on the interval. If you have the theorem that every polynomial is continuous at every x, you can use that. If not, look at [itex||f(x)- f(a)|= |(x- a)^2+ 2(x- a)|= |x- a||x- a+ 2|[/itex].
     
  10. Mar 2, 2012 #9
    Well, then how do I prove continuity on the interval?
     
  11. Mar 2, 2012 #10

    Mark44

    Staff: Mentor

    What is the statement of the problem? From your work it seems that you need to show that for some number c in [-3, 3], f(c) = 2. You don't show the problem statement, so I'm not sure that you need to prove that your function is continuous on that interval. Since f is a polynomial, and polynomials are continuous everywhere, it seems to me that your focus should be on proving that f(c) = 2 for some number in [-3, 3]. As HallsOfIvy already said, there's probably a theorem in you book about the continuity of polynomials.
     
    Last edited: Mar 2, 2012
  12. Mar 2, 2012 #11
    I don't have the statement of the problem; this was a bonus question on a quiz that she wrote on the whiteboard.
     
  13. Mar 2, 2012 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Your teacher wrote "also cont. (-3,3) - could use poly". I'd interpret this to mean that you should use the fact that polynomials are continuous functions to say that it's continuous, not prove it's continuous from scratch. The idea of this exercise isn't to prove continuity, it's to use the IVT.
     
  14. Mar 2, 2012 #13
    I believe that is all my teacher wanted me to do, to be able to use the Intermediate Value Theorem. So, what is missing from my own solution to the problem?
     
  15. Mar 2, 2012 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You didn't give a reason why f(x) is continuous. I think she just wanted you to say 'f is continuous on [-3,3] because f is a polynomial'.
     
  16. Mar 2, 2012 #15
    So, is every other part to my solution correct?
     
  17. Mar 2, 2012 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Pretty much. Given that you know f is continuous because it's a polynomial, you can omit the limit parts and just say what f(-3) and f(3) are. Then I think she also wants you to explicitly say how you used the IVT.
     
  18. Mar 3, 2012 #17
    Okay, so I found an intermediate value theorem question in my textbook. The question is verify that the intermediate value theorem applies to the indicated interval. My solution to the problem is in the attached file. Would this be an appropriate answer on a test?
     

    Attached Files:

  19. Mar 3, 2012 #18

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That looks pretty good to me. Except that f(0)=(-1). And the IVT doesn't find a value of x, it just tells you one exists, as you say later on.
     
  20. Mar 3, 2012 #19

    Mark44

    Staff: Mentor

    One other thing: you say "... the given interval, [], will be continuous..."

    It's the function, not the interval, whose continuity is being asserted.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Intermediate Value Theorem
Loading...