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Homework Help: Finding a solution

  1. Nov 23, 2008 #1
    Show that the equation xe^x = 1 has a solution between 0<x<1.

    Well for this I simply changed to x = 1/e^x

    Iterated starting with x=1, and got 0.567

    Is that the right way to go about this problem?

    Then it asks, prove that the solution is unique, which I guess means there is no other solution for the equation but 0.567.

    How would I go about this part?
  2. jcsd
  3. Nov 23, 2008 #2
    Ok perhaps wrong, another way:

    There is a solution where x0 is an element of [0,1]

    where f(x) = xe^x - 1

    f(0) = -1 < 0 < e^1 - 1 = f(1)

    By the intermediate value theroem there exists an x0 in the interval with f(x0) = 0.

    Still unsure on the second part, think it uses Rolle's theorem.
  4. Nov 23, 2008 #3
    Think I have it:

    For a,b in the interval, where a is not equal to b, then f(a) can not be equal to f(b) if x0 is unique.

    Then suppose f(a) does equal f(b), apply Rolles theorem:

    => there is an x0 in the interval where f'(x0) = 0

    But f'(x) = xe^x + e^x which can never be equal to zero since it is always positive for our interval [0,1] and all positive values of x, which is a contradiction.

    Now is the question asking me if the result is unique in the interval or for all x?

    Edit: So f'(x) is always posive for [-1,inf). Am i supposed to show the lim x-> -inf or something?
    Last edited: Nov 23, 2008
  5. Nov 24, 2008 #4


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    Science Advisor

    yes, what you have done is perfectly good.
  6. Nov 24, 2008 #5
    Is there a way to show that the solution is unique for all x? i can graph it and see it is, it's monotonic for [-1,inf), do i have to show for all (-inf,-1) that f(x) is always negative?
  7. Nov 24, 2008 #6


    Staff: Mentor

    Since you have shown that f(0) = -1, f(1) > 0, and f'(x) > for x in [0, 1], so f(x) = 0 for only a single value. A function whose derivative is positive on an interval can't cross the x-axis more than once on that interval.
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