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Linear Interpolation watch

1. 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.

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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?
2. (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.
3. (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)
4. The formula is
The formula is
This formula is awesome. And also, is this allowed for FP1 edexcel if you know by any chance lol
6. (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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Updated: January 5, 2013
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