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Finding range of f(x)

  1. Apr 12, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Let f:[0,1]→R be a function. Suppose the function f is twice differentiable, f(0)=f(1)=0 and satisfies [itex]f"(x)-2f'(x)+f(x)≥e^x[/itex], x:[0,1]. Then find range of f(x).


    3. The attempt at a solution
    Strictly speaking, this has been one of the toughest problems I've ever encountered throughout my course in Calculus. So, obviously I don't have an idea where to start with. But atleast, from the given information, I can say that f'(x) must be zero somewhere in [0,1] according to Rolle's Theorem. Since the inequation also involves a second order derivative, finding a solution of this differential inequation (never heard of diff. inequation btw) is beyond my reach. :cry:
     
  2. jcsd
  3. Apr 12, 2014 #2

    Mark44

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    If you turn your inequality into an equation, can you solve that?

    I did that, and got a solution that consists of a quadratic times e2; i.e., y(x) = (Ax2 + Bx + C)ex. The quadratic is a parabola that opens up and has x-intercepts at 0 and 1. My thinking is that if f(x) is any twice-differentiable function that is at least as large as y(x), and intersects the x-axis at 0 and 1, then that will be a solution of your differential inequality.

    I'm not 100% confident in my reasoning here - I haven't ever seen differential inequalities, either. If the direction I took makes sense, you could at least put a lower bound on the range.
     
  4. Apr 12, 2014 #3

    haruspex

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    I'm not sure what the question is asking. Clearly f is not fully defined, so different functions satisfying the condition might have different ranges. Is it the union of the ranges that's required? The set of ranges?
     
  5. Apr 12, 2014 #4
    Well, the range is just R. No? Unless you are talking about the image of f. By the way, I do not believe that f is uniquely defined, as haruspex pointed out.
     
  6. Apr 12, 2014 #5

    PeroK

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    Hint: Let ##g(x) = f(x)e^{-x}##
     
  7. Apr 12, 2014 #6

    haruspex

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    Why? I think it unlikely that arbitrarily negative values can be reached.
    If you take the inequality as exact, as Mark44 did, and plug in the known facts, you'll find 2a >= 1, b = -a and c = 0. You can then find a local min in the interval as a function of a.
     
  8. Apr 12, 2014 #7

    PeroK

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    Let ##g(x) = f(x)e^{-x}##

    g satisfies g(0) = g(1) = 0 and g''(x) ≥ 1

    Let ##g_n(x) = nx(x-1)##

    So: ##f_n(x) = nx(x-1)e^x##

    So, the range of f is <= 0 and can be arbitrarily large and -ve.
     
  9. Apr 12, 2014 #8

    utkarshakash

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    Ah, thanks for pointing out. The question actually gives me 4 options and asks me to select a correct one. Here are those 4 options:-

    Q. Which of the following is true for 0<x<1?
    A) 0<f(x)<∞
    B)-1/2 < f(x)< 1/2
    C)-1/4<f(x)<1
    D)-∞<f(x)<0
     
  10. Apr 15, 2014 #9

    utkarshakash

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    Did anyone get the answer?
     
  11. Apr 15, 2014 #10

    haruspex

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    PeroK's analysis convinced me.
     
  12. Apr 15, 2014 #11

    HallsofIvy

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    You two are using different definitions of "range". crownedbishop is saying that if f is "from A to B" then its range is "B", even if f(A), the "image" of f (or, better, the "image of A under f")is a proper subset of B.
     
  13. Apr 15, 2014 #12

    haruspex

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    True, range can mean either codomain or image of domain. But it seems clear that image of domain is what's wanted here.
     
  14. Apr 15, 2014 #13

    utkarshakash

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    Your approach is a perfect one. May I know how did it occur to you to assume another function g(x)=f(x)e^-x? Were you led by your intuition or just a lucky guess? How exactly did you figure it out? Were you able to realize it quickly or did it took a while?
     
  15. Apr 15, 2014 #14

    PeroK

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    I just saw the pattern of f'' - 2f' + f and thought I might be able to find a function whose second derivative that was. I used what's called an "integrating factor", a technique used in solving of differential equations:

    Since you ask, it was the first thing I thought of when I saw the question.
     
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