... consider the function between -3 and -2 I can see I change of sign on the function from -ve to +ve. -6 -183 -5 -97 -4 -41 -3 -9 -2 5 -1 7 0 3 1 -1 To home in on a more accurate answer I need to investigate the decimal places between -3 and -2 x f(x) -3 -9 -2.9 -6.889 -2.8 -4.952 -2.7 -3.183 -2.6 -1.576 -2.5 -0.125 -2.4 1.176 -2.3 2.333 -2.2 3.352 -2.1 4.239 -2 5 This method is repeated until a sufficient number of decimal places are achieved. In this investigation I think 5 decimal places will be enough -2.5 -0.125 -2.49 0.011751 -2.48 0.147008 -2.47 0.280777 -2.46 0.413064 -2.45 0.543875 -2.44 0.673216 -2.43 0.801093 -2.42 0.927512 -2.41 1.052479 -2.4 1.176 -2.5 -0.125 -2.499 -0.11126 -2.498 -0.09753 -2.497 -0.08382 -2.496 -0.07012 -2.495 -0.05644 -2.494 -0.04277 -2.493 -0.02912 -2.492 -0.01548 -2.491 -0.00186 -2.49 0.011751 From this I can see that the route lies between -2.491 and -2.490±0.0005 and is found at 0.011751 ± 0.0000005. This method works for most functions of x but can still cause anomalous results in the occasion of certain instances where the method will not work as there being two routes within an interval. When I plot a graph with this characteristic both routes are clearly distinguishable visually, however when I find the solutions numerically it only shows one change of sign being the first one and the second ...

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