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Compound functions

  1. May 1, 2007 #1
    1. The problem statement, all variables and given/known data
    Sketch sin (1/x), x sin (1/x) and x^2 sin (1/x) and show they are discontinuous at x=0.

    3. The attempt at a solution

    I *know* how to do this problem with the usual curve sketching methods (important points, asymptotes, turning points, "tends to") but my book has a method that involves breaking the compound function into smaller ones. I don't know how to break x sin (1/x) into compound functions, though. I don't think it's necessary to totally break it down anyway since the graphs all have about the same shape.

    I can do sin (1/x)... I think... do check my solution too.

    y = f(x) = sin (1/x)
    w = g(x) = 1/x
    u = h(x) = sin w

    sin w can reach 1 when w = π/2, 5π/2... and -1 when w = 3π/2, 7π/2...
    and is 0 for any integral multiple of π.

    Substituting for x, sin (1/x) reaches

    1 when x = 2/π, 2/5π...
    -1 when x = 2/3π, 2/7π...
    0 for x = 1/kπ and k is any integer.

    Deduce that f(x) tends to slightly below 0 at negative infinity and slightly above 0 at positive infinity.

    Also substitute values for x = 1/kπ: 1/π, 1/2π, 1/3π, 1/4π

    When plotting a graph the intervals between successive 0s become smaller to negligible. Formula for 1, -1 also have similar form, so values for -1, 1 and 0 cluster as x tends towards 0 (or k tends towards infinity). By inspection the values oscillate between 1, 0 and -1, and that describes the graph of sin (1/x). Graph is discontinuous for 1/x at x=0.

    I could do the rest using calculus, but I'm told there may be no need to (or told this so that I have to do everything the long way). The values for 0 should be at the same value for x in all 3 equations. I *know* that the y values of the turning points are not the same, but that's after differentiation.
     
    Last edited: May 1, 2007
  2. jcsd
  3. May 1, 2007 #2
    Calculus method: differentiating the latter two expressions gives is

    f(x) = x sin (1/x)
    f'(x) = sin(1/x) - (1/x) cos(1/x)

    g(x) = x^2 sin (1/x)
    g'(x) = 2x sin(1/x) - cos(1/x)

    when f'(x), g'(x) = 0

    tan(1/x)=1/x for f(x)
    tan(1/x)=1/2x for g(x)

    We can get these values at our own leisure (isn't this fun...) by substituting for 1/x, eg

    h(v) = tan v - v
    j(u) = tan u - 0.5v

    and applying the Newton-Raphson iterative method. That's pretty tedious, considering that the function oscillates an infinite number of times as it tends to 0, which is why I'm seeking a non-calculus solution. I don't *need* to know where the points are specifically. I've deduced graph shapes and the coordinates of all points where y=0, the final bit is how the turning point coordinates vary.
     
  4. May 1, 2007 #3

    Dick

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    If you've deduced graph shapes you should know x*sin(1/x) and x^2*sin(1/x) are continuous at 0, aside from the technicality that they aren't defined there.
     
  5. May 2, 2007 #4
    I know that, Dick... all 3 equations have the term sin (1/x), which is undefined at 0. But I also have to sketch the three curves, of which I have done one. I know the points where y=0 for all three curves, and how they look like, but the problem is the the turning points.
     
  6. May 2, 2007 #5

    Dick

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    I think you only need a rough sketch of the function. No need to compute exact turning points etc. There are a infinite number of them...
     
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