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Intermediate Value Theorem

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Use the IVT to find an interval of length 1/2 containing a root of
    f(x)=x3+ 2x + 1

    2. Relevant equations

    Intermediate Value Theorem: If f(x) is continuous on a closed interval [a, b] and f(a)≠f(b) then for every value M between f(a) and f(b) there exists at least one value c[itex]\in[/itex](a, b) such that f(c) = M

    3. The attempt at a solution

    So I am thinking with this what I need to do is take any 1/2 length interval and plug in those values for x. I took [0, 1/2] and plugged it in. I got

    f(0) = 1

    f(1/2) = 2.125 or 2 1/8. It just asked to find an interval. So I would think I could say f(c) exists somewhere between f(0) and f(1/2) because they are both continuous functions.

    Let me know if this is right, because the math tutor told me it was wrong, and I think he's wrong.
     
  2. jcsd
  3. Apr 15, 2013 #2

    LCKurtz

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    You are looking for a root, that is where f(c) = 0. So it would be good if your interval had f(x) having opposite signs at the ends.
     
  4. Apr 15, 2013 #3
    root

    I can't even remember roots. I guess I have to brush up on that.

    By the way, just telling me that with that ONE SENTENCE was more than anything that particular tutor could do. I appreciate your succinctness.

    Edit:

    Ok, so to solve this, I could choose [-1/2, 0] ?

    [-1/8, 1]

    Zero exists between those two points, its a continuous function, so f(c)= 0 exists. Look alright?
     
    Last edited: Apr 15, 2013
  5. Apr 15, 2013 #4

    LCKurtz

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    Yes, but you could phrase it better: "f(-1/2)=-1/8 and f(0) = 1 so by the intermediate value theorem there exists a c between -1/8 and 1 such that f(c) = 0".
     
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