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# C3 - Linear Interpolation/Decimal Search Help watch

1. 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 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!
2. anyone
3. 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 :\
4. Linear Interpolation is the method that involves two similar triangles. Sketch a diagram.
5. 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.
6. 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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Updated: October 21, 2008
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