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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