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 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=___________?

Asked on by bilemem

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embizze | High School Teacher | (Level 1) Educator Emeritus

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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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