# `e^(arctan(x)) = sqrt(x^3 + 1)` Use Newton's method to find all roots of the equation correct to eight decimal places.

### Textbook Question

Chapter 4, 4.8 - Problem 28 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to remember how Newton's method is used, to evaluate the roots of a transcendent equation.

`x_(n+1) = x_n - (f(x_n))/(f'(x_n))` , where `x_n` is the approximate solution to the equation f(x) = 0.

You need to solve for x the equation `e^(arctan x) = sqrt(x^3+1)` , hence, you need to have the function in the form `f(x) = 0` , such that:

`e^(arctan x) - sqrt(x^3+1) = 0`

Notice that for x = 0 yields:

`e^(arctan 0) - sqrt(0+1) = e^0 - sqrt 1 = 1-1 = 0`

You may consider the first approximation:

`x_1 = 0 - (f(0))/(f'(0))`

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

`(f'(0)) = (e^(arctan 0))/(1+0^2) - (0^2)/(2sqrt(0^3+1)) = 1 !+0`

`x_1 = 0 - 0/1 = 0`

Hence, evaluating the root of the given transcendent equation, using Newton's method, yields x = 0.