# Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your answer to four decimal places.)  (1/3)x^3 + (1/2)x^2 + 3=0  ,  x1 = −3

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Notice that the problem provides the initial approximation `x_1=-3`  such that:

`l(x_2) = f(x_1) + f'(x_1)(x_2-x_1)`

`l(x_2) = 0 =gt f(x_1) + f'(x_1)(x_2-x_1) = 0`

`(x_2-x_1) =(-f(x_1)) / (f'(x_1) )`

`x_2 = x_1 - (f(x_1)) / (f'(x_1) )`

You need to evaluate `f(x_1)` ...

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