... interval, 1.25. f(1.25) = 2.016, so f(1.25)> 0. Since f(1) <0, the root is in [1, 1.25]. The midpoint of this reduced interval is 1.125. f(1.125) = 0.244, so f(1.125)> 0. Since f(1) <0, the root is in [1, 1.125]. The method then continues in this manner until the required degree of accuracy is achieved. However, this takes a long time to converge the root, and can be solved more rapidly using a spreadsheet. To put my data into a spreadsheet, I first needed to design one that achieved the desired purpose: finding the roots of the equation using the change of sign method. Below, the formula which I typed into my spreadsheet are shown: Change of sign method y=5x^3-7x+1 a b (a+b)/2 f(a) f(b) f((a+b)/2) 1 2 =(A5+B5)/2 =5*A5^3-7*A5+1 =5*B5^3-7*B5+1 =5*C5^3-7*C5+1 =IF(D5*F5<0,A5,C5) =IF(D5*F5<0,C5,B5) =(A6+B6)/2 =5*A6^3-7*A6+1 =5*B6^3-7*B6+1 =5*C6^3-7*C6+1 =IF(D6*F6<0,A6,C6) =IF(D6*F6<0,C6,B6) =(A7+B7)/2 =5*A7^3-7*A7+1 =5*B7^3-7*B7+1 =5*C7^3-7*C7+1 =IF(D7*F7<0,A7,C7) =IF(D7*F7<0,C7,B7) =(A8+B8)/2 =5*A8^3-7*A8+1 =5*B8^3-7*B8+1 =5*C8^3-7*C8+1 =IF(D8*F8<0,A8,C8) =IF(D8*F8<0,C8,B8) =(A9+B9)/2 =5*A9^3-7*A9+1 =5*B9^3-7*B9+1 =5*C9^3-7*C9+1 =IF(D9*F9<0,A9,C9) =IF(D9*F9<0,C9,B9) =(A10+B10)/2 =5*A10^3-7*A10+1 =5*B10^3-7*B10+1 =5*C10^3-7*C10+1 From this spread sheet I can find the upper and lower bounds of the intervals so that I can narrow down the accuracy of the root of the equation. Shown above are only ...

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