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 Problem: secant method 2x^3-2x-5, a=1, b=4 [ Calculator, Method and examples ]Solution:Your problem -> secant method 2x^3-2x-5, a=1, b=4Here 2x^3-2x-5=0Let f(x) = 2x^3-2x-51^(st) iteration :x_0 = 1 and x_1 = 4f(x_0) = f(1) = -5 and f(x_1) = f(4) = 115:. x_2 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))x_2 = 1 - (-5) * (4 - 1)/(115 - (-5))x_2 = 1.125:. f(x_2)=f(1.125)=2*(1.125)^3-2*(1.125)-5=-4.40232^(nd) iteration :x_1 = 4 and x_2 = 1.125f(x_1) = f(4) = 115 and f(x_2) = f(1.125) = -4.4023:. x_3 = x_1 - f(x_1) * (x_2 - x_1)/(f(x_2) - f(x_1))x_3 = 4 - 115 * (1.125 - 4)/(-4.4023 - 115)x_3 = 1.231:. f(x_3)=f(1.231)=2*(1.231)^3-2*(1.231)-5=-3.73123^(rd) iteration :x_2 = 1.125 and x_3 = 1.231f(x_2) = f(1.125) = -4.4023 and f(x_3) = f(1.231) = -3.7312:. x_4 = x_2 - f(x_2) * (x_3 - x_2)/(f(x_3) - f(x_2))x_4 = 1.125 - (-4.4023) * (1.231 - 1.125)/(-3.7312 - (-4.4023))x_4 = 1.8203:. f(x_4)=f(1.8203)=2*(1.8203)^3-2*(1.8203)-5=3.42224^(th) iteration :x_3 = 1.231 and x_4 = 1.8203f(x_3) = f(1.231) = -3.7312 and f(x_4) = f(1.8203) = 3.4222:. x_5 = x_3 - f(x_3) * (x_4 - x_3)/(f(x_4) - f(x_3))

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