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C3 - Linear Interpolation/Decimal Search Help watch

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    I really cannot understand the method used to use linear interpolation, my teacher is a waste.

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    q) Show that a root exists x=a of the equation x^3-6x+3=0 such that 2 < a < 3. Use decimal search to find an interval of width of width 0.001 in which this root lies.

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    I've subbed in 2 and 3 into the function, to get the root is between (2,-1) and (3,12)

    after that I'm pretty clueless on how to do this with linear interpolation as i don't really know the method.. if anyone could tell me what it is, much appreciated!
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    anyone :confused:
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    What exam board are you on?? I'm with AQA and this is on FP1 for us.

    Anyway, yeah sub 2 and 3 in to get 2 values. If the sign changes then there's a root between those two values. Then what we normally do is find the middle value and sub that in, to find a closer 2 values the root is between, then keep on going. But we normally stop at 0.25 interval width.. I don't really know how to do it with a 0.001 width :\
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    Linear Interpolation is the method that involves two similar triangles. Sketch a diagram.
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    Ah.. its OCR and my teacher went through some stuff with fractions and things, not just set intervals. Ah I dunno. - I mean if this was iteration then it'd be fine.. but this method takes the mick.
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    Ah yes, Sorry I was getting muggled up, sorry.
    find the x and y co-ords. then find the gradient. form the equation of the straight line that connects the 2 points of the curve together. That gives you a root that's close to the root of the curve. Remember to set y=o to find the x value of the root of the straight line. Interval width 0.001 will simply mean to 3 decimal places.

    That's what I have in my further maths notes for it anyway. I dunno what that guy was on about triangles...
 
 
 

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