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5 years since my last math class

  1. Sep 8, 2004 #1
    It has been at least 5 years since my last math class and I am kinda struggling.

    The question is:
    Use the intermediate value theorem to show that there is a root of ln x = e^-x on the interval (1,2). Find the correct root to 3 decimal values

    What i have so far
    f(1) = ln 1 - e^-1 = 0 which -.368< 0
    f(2) = ln 2 - e^-2 = 0 which .558 > 0

    am I doing this correct or am I really off base ?

    thanks joe
  2. jcsd
  3. Sep 8, 2004 #2

    matt grime

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    how can log(1) -e^{-1}=0? it doesn't why is that equals sign there? but apart from that you are in the right ball park.

    f(1) is negative,
    f(2) is positive, hence since the function is continuos it must pass through zero at some intermediate value of x between 1 and 2.

    finding the root is a different matter. many methods exist, what do you know about?
  4. Sep 8, 2004 #3
    I don't know much about it , and reading the scetion is difficult to understand. I have been going to math lab and looking for a tutor to assist me with my struggles. but at least i am on the right track.

    You said that this function passes through 0. Do I need to pick a point greater than one but less than two and plug back into the function

  5. Sep 8, 2004 #4
    yes sort of !

    Specifically u need some good iteration method now to finish the work,
    two methods you can use now is bisection method and regula-falsi or the false position method.

    Ofcourse there are other methods ... but these are fairly easy to implement.

    -- AI
  6. Sep 9, 2004 #5


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    Use the intermediate value theorem to show that there is a root of ln x = e^-x on the interval (1,2). Find the correct root to 3 decimal values

    Let f(x)= ln x- e-x so that ln x= e-x is the same as f(x)= 0.
    (and f is continuous for x> 0)

    f(1)= ln(1)- e-1= -e-1< 0 (your "= 0" was left over from the equation and doesn't apply for x= 1).
    f(2)= ln(2)- e-2= 0.558> 0.

    The intermediate value theorem (one version) says: if f(x) is continuous on [a,b], f(a)< 0 and f(b)> 0 then there exist c in [a,b] such that f(c)= 0.

    Since for we have f(1)<0, f(2)> 0, there exist c in [1, 2] such that f(c)= 0 or
    ln(c)= e-c.

    As TenaliRaman said, one good way of finding that number is "bisection" and I suspect that is what is intended specifically because it uses the intermediate value theorem.

    We know, from the intermediate value theorem that there is a solution in [1, 2]. Where? No idea! So try the midpoint just because it is easiest: midpoint of [1, 2] is x= 1.5. ln(1.5)- exp(-1.5)= 0.182> 0. That's not a solution since it is positive but we now know (again by the intermediat value theorem) that there is a solution between 1 and 1.5. Again, try the midpoint: (1+ 1.5)/2= 1.25. Determine whether ln(1.25)- e-1.25 is positive or negative to decide whether there is a solution between 1 and 1.25 or between 1.25 and 1.5 and take the midpoint of that interval. Continue until you get 3 decimal places right.
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