Use Newton\'s method with the specified initial approximation x1 to find x3 1/3

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Explanation / Answer

(1/3) * x^3 + (1/2) * x^2 + 5 = 0 ======> f (x) = (1/3) * x^3 + (1/2) * x^2 + 5 f '(x) = x^2 + x lets start with X1 = -3 Xn+1 = Xn - [ f(x) / f '(x) ] Xn+1 = Xn - [ ((1/3) * X^3 + (1/2) * X^2 + 5) / (X^2 + X) ] ======> X1 = -3 X 2 = -3 - [ ((1/3) * (-3)^3 + (1/2) * (-3)^2 + 5) / ( (-3)^2 + (-3) ] ====> X2 = -3.083333333 X 3 = -3.083333333 - [ ((1/3) * (-3.083333333)^3 + (1/2) * (-3.083333333)^2 + 5) / ( (-3.083333333)^2 + (-3.083333333) ] ====> X3 = -3.080600601 X 4 = -3.080600601 - [ ((1/3) * (-3.080600601)^3 + (1/2) * (-3.080600601)^2 + 5) / ( (-3.080600601)^2 + (-3.080600601) ] ====> X4 = -3.080597592 X 5 = -3.080597592 - [ ((1/3) * (-3.080597592)^3 + (1/2) * (-3.080597592)^2 + 5) / ( (-3.080597592)^2 + (-3.080597592) ] ====> X5 = -3.080597592 The root is -3.08060

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