# Use Newton's method with the specified initial approximationUse Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation....

Use Newton's method with the specified initial approximation

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 +10 =0,

x1=-3

x3=___________?

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### 1 Answer

Given `f(x)=1/3 x^3+1/2x^2+10`

`f'(x)=x^2+x`

Newtons method begins with a "guess", `x_1` , then proceeds towards the actual value using `x_n=x_(n-1)-(f(x_n))/(f'(x_n))`

`x_1=-3`

`x_2=-3-5.5/6=-3.91667`

`x_3=-3.91667-(-2.3475)/(11.42363)=-3.71117`

**To 4 decimal places we have -3.71117**

Compare to -3.696058 correct to 5 decimal places.