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=___________?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.

See
This Answer Now

Start your subscription to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your Subscription

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.

Approved by eNotes Editorial Team