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    f(x) = 3tan(x/2) - x - 1 ... x is between plus minus pi

    Given that f(x) = 0 has a root between 1 and 2, use linear interpolation once on the interval [1, 2] to find an approximation to this root.

    ----
    How to do this? I did f(1) and f(2) to form:

    (x-1)/0.3601... = (2-x)/1.672...

    but this gives x as 0.47... which is clearly wrong. Is there a different approach to it being in radians, etc?
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    (Original post by Lunch_Box)
    f(x) = 3tan(x/2) - x - 1 ... x is between plus minus pi

    Given that f(x) = 0 has a root between 1 and 2, use linear interpolation once on the interval [1, 2] to find an approximation to this root.

    ----
    How to do this? I did f(1) and f(2) to form:

    (x-1)/0.3601... = (2-x)/1.672...

    but this gives x as 0.47... which is clearly wrong. Is there a different approach to it being in radians, etc?
    It has to be in radians. I can't really tell what you're doing since you've only shown 1 line of working, but your first iteration should be 0=f(1)+[f(2)-f(1)]*(x-1)/(2-1), after which you solve for x.
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    (Original post by valhalla92)
    It has to be in radians. I can't really tell what you're doing since you've only shown 1 line of working, but your first iteration should be 0=f(1)+[f(2)-f(1)]*(x-1)/(2-1), after which you solve for x.
    I'm unaware of that formula...

    I use this method of similar triangles:




    I have done:

    (x-1) / f(1) = (2 - x) / f(2)
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    The formula is \frac{af(b)-bf(a)}{f(b)-f(a)}
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    (Original post by Damask-)
    The formula is \frac{af(b)-bf(a)}{f(b)-f(a)}
    This formula is awesome. And also, is this allowed for FP1 edexcel if you know by any chance lol
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    (Original post by Lunch_Box)
    So I have to take into account the negatives? And also, is this allowed for FP1 edexcel if you know by any chance lol
    Yes you do, and it's a perfectly valid way to go about solving the question, so you can use it in FP1. Our teacher got us to derive it in a lesson so we'd understand where it comes from.
 
 
 
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