# Use Newton's method with the specified initial approximation 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 +10 =0, x1=-3 x3=___________?

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.

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.

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