MHB Sydney's question at Yahoo Answers regarding root approximation

[MarkFL]

Here is the question:

CALCULUS HELP PLEASE!?

Approximate the solution to the following equation that satisfies the given condition: x^(4) + x - 3 = 0 ; x is greater than or equal to 0Please help I'm having so much trouble!

Here is a link to the question:

CALCULUS HELP PLEASE!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Sydney,

First, let's take a look at a plot of the function:

View attachment 582

We see the positive root is near $x=1$, so that will be a good initial guess.

Newton's method gives us the recursion:

$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

In our case, this is:

$\displaystyle x_{n+1}=x_n-\frac{x_n^4+x_n-3}{4x_n^3+1}=\frac{3(x_n^4+1)}{4x_n^3+1}$

where $x_0=1$.

Now, I have a TI-89 Titanium graphing calculator, so what I do is enter the following:

1 [ENTER] Result: 1
(3(ans(1)^4+1))/(4ans(1)^3+1) ♦[ENTER] Result: 1.2
♦[ENTER] Result: 1.16541961577
♦[ENTER] Result: 1.16403726916
♦[ENTER] Result: 1.16403514029
♦[ENTER] Result: 1.16403514029

Since the last two approximations are the same, we have exceeded the limit of accuracy of the calculator, and we know the required root, to 12 digits, is:

$x\approx1.16403514029$