# Find the root of the equation. Give your answers correct to six decimal places?

##
#x^3-x=2#

(a) Use Newton's method with x1 = 1.

(b) Solve the equation using x1 = 0.6 as the initial approximation.

(c) Solve the equation using x1 = 0.58. (You definitely need a programmable calculator for this part.)

(a) Use Newton's method with x1 = 1.

(b) Solve the equation using x1 = 0.6 as the initial approximation.

(c) Solve the equation using x1 = 0.58. (You definitely need a programmable calculator for this part.)

The solution is

We have:

Let

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

# :. x_(n+1) = x_n - (x_n^3-x_n-2) / (3x_n^2-1) #

(a) If we start with

So we see that very rapidly the Newton-Rhapson method converges to the solution

.

(b) If we start with

So again we see that the Newton-Rhapson method converges to the solution

(c) If we start with

So again we see that the Newton-Rhapson method converges to the solution

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Sure, please provide the equation for which you'd like to find the root.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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