... will continue with the decimal search, but now using increments of 0.01 within the interval [0.8,0.9]. x 0.80 0.81 0.82 0.83 0.84 f(x) -0.184 -0.138727 -0.092096 -0.044089 0.005312 This indicates that the root lies in the new interval [0.83,0.84]. I will continue the process using the increments of 0.001 within the interval [0.83,0.84]. x 0.830 0.831 0.832 0.833 0.834 0.835 f(x) -0.044089 -0.039211927 -0.034320896 -0.029415889 -0.024496888 -0.019563875 x 0.836 0.837 0.838 0.839 f(x) -0.014616832 -0.009655741 -0.004680584 0.000308657 The root therefore lies in the interval [0.838,0.839]. I shall now use increments of 0.0001 in the interval [0.838,0.839]. x 0.8380 0.8381 0.8382 0.8383 0.8384 0.8385 f(x) -0.004680584 -0.00418229398 -0.0036838631 -0.00318529134 -0.00268657869 -0.00218772512 x 0.8386 0.8387 0.8388 0.8389 0.8390 f(x) -0.00168873063 -0.00118959519 -0.00069031878 -0.00019090139 0.000308657 Therefore the root lies between 0.8389 and 0.8390. To three decimal places, the root = 0.839. Error Bounds A change of sign method such as the one used, provides bounds within which a root lies so that the maximum possible error in a result is known. When x = 0.8385, f(0.8385) = -0.00218772512 When x = 0.8395, f(0.8395) = 0.00280856463 The error bounds of the root 0.839 are 0.839 ± 0.0005. However, I am able to say that I have a more accurate solution, as I know that the root lies in the interval [0.8389,0.8390]. Failure ...

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