Use Newtons method with the specified initial approximation x1 to find x3, the third approximation to the
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5. x + 2x - 4 - 0, x = 1 6. x - x - 1= 0, x - 1 7. x* - 20 = 0, x = 2 8. x* + 2 = 0, x = -1
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5 fx x 2x 4 I I2 1 Now I 12 4 312 2 1 1 Now 1 2 6 fx xx 1 fx 3x 2x so n1 In x2 1 fx ...View the full answer
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