# 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 +...

Use Newton's method with the specified initial approximation *x*1 to find *x*3, 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)` such that:

`f(x_1) = f(-3) = -27/3 + 9/2 + 3 = -6 + 9/2 `

`f(-3) = -3/2`

`f'(x) = x^2 + x =gt f(-3) = 9 - 3 = 6`

`x_2 = -3 + 3/12 =gt x_2 = -3 + 1/4 =gt x_2 = -11/4`

`x_3 = x_2 - (f(x_2)) / (f'(x_2) )`

`f(-11/4) = -(1/3)(1331/64) + 121/32 + 3`

`f(-11/4) = (-1331 + 726 + 576)/192`

`f(-11/4) = -29/192`

`f'(-11/4) = 121/16 - 11/4 = (121-44)/16 = 77/16`

`x_3 = -11/4 + (29/192)/(77/16)`

`x_3 = -2.75 + 464/14784`

`x_3 = -2.75 + 0.03`

`x_3 = -2.71`

**Hence, evaluating the roots `x_2 ` and `x_3` using Newton's method yields `x_2 = -2.75 ; x_3 = -2.71` .**