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Calc 1 proof IVT Rolles theorem

  1. Oct 24, 2014 #1

    462chevelle

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    Gold Member

    1. The problem statement, all variables and given/known data
    Use the intermediate value theorem and rolles theorem to prove that the equation has exactly one real solution
    2x-2-cos(x)=0



    3. The attempt at a solution
    Let the interval be [a,b] and let f(a)<0 and f(b)>0
    Then by the IVT there must be at least one zero between a and b.
    f'(x)=2+sin(x)
    since f'(x) doesnt = 0 anywhere and its always >0, therefore f(x) is increasing throughout its entire domain. Therefore f(a) cannot = f(b) anywhere.

    I feel like im doing a bad job at explaining this, but this is my first proof for class ever, other than geometry in highschool and i was bad at it. Is there anything terribly wrong or that could be improved upon at all?
     
    Last edited by a moderator: Oct 24, 2014
  2. jcsd
  3. Oct 24, 2014 #2

    Mark44

    Staff: Mentor

    No, you can't do this. You need to find numbers a and b for which the two inequalities are true. Just try a few values to see if you can get the function to change sign going from one number to the other.
     
    Last edited: Oct 24, 2014
  4. Oct 24, 2014 #3

    462chevelle

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    Gold Member

    My confusion is that there is no points where f(a)=f(b) so i dont really know what to do with rolles theorem. Ill rewrite that part with some numbers.
     
  5. Oct 24, 2014 #4

    Mark44

    Staff: Mentor

    Let me change what I said. Use the IVT first. That will show that there is at least one root.
     
  6. Oct 24, 2014 #5

    462chevelle

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    alright, so if i make the interval [0,2] i get a - then a +. so that would satisfy the IVT. then i use my info that the derivative is always positive so the function is always increasing, therefore only crosses the x axis once. How do i factor in rolles theorem? or would i have to use contradiction or something? Or could i say, that since there is no interval where f(a)=f(b) then if the function is increasing in one point then it must be increasing throughout the entire domain.
     
  7. Oct 24, 2014 #6

    Mark44

    Staff: Mentor

    I think you have to use contradiction. From the IVT, you know that there is a zero (around 1.2, BTW). Call this a.

    Now suppose that there is another solution b ≠ a, such that f(b) = 0.
    What does Rolle's say?
    What do you know about f'(x)?
     
  8. Oct 24, 2014 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You have already said
    So if f(a)= 0 then we cannot have f(b)= 0 for any other b. that is, f(x)= 0 has at most one solution.

    As others have said you cannot simply assert
    until you know that there exist values of x where f(x)< 0 and where f(x)> 0.

    You are given that f(x)= 2x- 2- cos(x). Further, you know that cos(x) has value only between -1 and 1. Suppose x is some very large, negative, number, say x= -1000000. What can you say about f(x)? Suppose x is some very large, positive, number, say x= 1000000. What can you say about f(x)?
     
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