`x/(x^2 + 1) = sqrt(1 - x)` Use Newton's method to find all roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations.

Textbook Question

Chapter 4, 4.8 - Problem 25 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`x/(x^2+1)=sqrt(1-x)`

`f(x)=x/(x^2+1)-sqrt(1-x)=0`

`f'(x)=((x^2+1)-x(2x))/(x^2+1)^2-(1/2)(1-x)^(-1/2)(-1)`

`f'(x)=(1-x^2)/(x^2+1)^2+1/(2sqrt(1-x))`

See the attached graph. From the graph the curve intersects x-axis at `~~` 0.8

`x_(n+1)=x_n-(x_n/((x_n)^2+1)-sqrt(1-x_n))/((1-(x_n)^2)/((x_n)^2+1)^2+1/(2sqrt(1-x_n)))`

Now x_1 `~~` 0.8

`x_2=0.8-(0.8/(0.8^2+1)-sqrt(1-0.8))/((1-0.8^2)/(0.8^2+1)^2+1/(2sqrt(1-0.8)))`

`x_2~~0.76757581`

approximate until eight decimal places are same

`x_3~~0.76682610`

`x_4~~0.76682579`

`x_5~~0.76682579`

To eight decimal places the root of the equation is `~~` 0.76682579

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