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Newton Rhapson + Graph Sketch

  1. Feb 1, 2006 #1
    "By sketching appropriate graphs (one of them a straight line) show that the function
    f(x) = 2sin(x) - 3x + 2

    has only one real root."

    this one confuses me from the word go. how can you sketch a straight line of that function when it has sin as part of it?
     
  2. jcsd
  3. Feb 1, 2006 #2

    HallsofIvy

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    The problem did say "graphs" (plural).
    How about y= 2 sin(x) and y= 3x- 2?

    Where the graphs cross, 2 sin(x)= 3x- 2 so 3 sin(x)- 3x+ 2= 0.

    (In the interest of beating a dead horse)
    Was that the exact wording of the problem? A function does not have "roots". A function has "zeroes", which are "roots" of the equation f(x)= 0.

    If I saw something like "find the roots of f(x)= x2" I would think "x2= 1 or x2= 4 or x2= 1000...???"
     
    Last edited: Feb 1, 2006
  4. Feb 1, 2006 #3
    yeah thats a word for word quote.

    ok so i set the two functions up and compared them and they cross once at approx 1.3, 1.9

    i thought that the 1.3 was the value that has to be used in the newton rhapson method, but the first iteration derivates so heavily from the origonal (i ended up with -0.49908) that i think im wrong.

    EDIT: Yeah somethings wrong...the more iterations i do the more the number jumps around.

    DOUBLE EDIT: Calculator was in wrong mode, im an idiot.
     
    Last edited: Feb 1, 2006
  5. Feb 1, 2006 #4

    HallsofIvy

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    Well, I wasn't going to mention it---
     
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