# `x^3 = tan^-1 (x)` Use Newton's method to find all roots of the equation correct to six decimal places.

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`x^3=tan^-1(x)`

`f(x)=x^3-tan^-1(x)=0`

`f'(x)=3x^2-(1/(x^2+1))`

To solve using Newton's method apply the formula,

`x_(n+1)=x_n-f(x_n)/(f'(x_n))`

Plug in f(x) and f'(x) in the formula

`x_(n+1)=x_n-((x_n)^3-tan^-1(x_n))/(3(x_n)^2-(1/((x_n)^2+1)))`

See the attached graph for finding the initial values of x_1.

The curve of the function intersects the x axis at `~~` 0.9 and -0.9

Let's solve for the first zero x_1=0.9,

`x_2=0.9-(0.9^3-tan^-1(0.9))/(3(0.9)^2-(1/(0.9^2+1)))`

`x_2~~0.90203199`

`x_3~~0.90202549`

`x_4~~0.90202549`

Now let's solve for the second zero , x_1=-0.9

`x_2=-0.9-((-0.9)^3-tan^-1(-0.9))/(3(-0.9)^2-(1/((-0.9)^2+1)))`

`x_2~~-0.9020443393`

`x_3~~-0.902025494`

`x_4~~-0.9020254924`

`x_5~~-0.9020254924`

So the roots of the equation are **0.902025 and -0.902025**