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I'm trying to use the Method of false position on the functions below to find solutions to within $10^{-5}$

(1) $f(x) = \ln (x-1) + \cos(x-1) = 0$, for $1.3 \leq x \leq2$


(2) $2x \cos(2x) -(x-2)^2 = 0$, for $[2,3]$ and $[3,4]$

I am trying to do this without using a computer program, and am stuck. Thanks

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Concerning function 1, what are you asking here different from this question of yours? – Américo Tavares Feb 4 '12 at 18:23
They are using different methods – James R. Feb 6 '12 at 10:21
Thank you! I had wrongly thought it was the same method, just different names. – Américo Tavares Feb 6 '12 at 12:48
up vote 1 down vote accepted
  1. Evaluate $f(1.3)$ and $f(2)$. (You will want to use a calculator.)
  2. Now, given just these two points, suppose $f$ is like a straight line through these two points. Where (i.e. for what $x$) would $f(x)=0$? (The $x$ you find this way, by pretending $f$ is a straight line inside the interval, is the false position.)
  3. Evaluate $f(x)$ for this $x$ (using the correct $f$ again now).
  4. You now have a smaller interval (either $[1.3,x]$ or $[x,2]$ -- whichever one contains a sign change for $f$) which you know contains the solution to $f(x)=0$.
  5. Go back to step 1, but now use your smaller interval instead of $[1.3,2]$.
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