Answer:
- x ≈ 0.760
- f(x) = 3x^4 -1
- [tex]x_{n+1}=\dfrac{9x^4+1}{12x^3}[/tex]
Step-by-step explanation:
Newton's method seeks a solution to the equation f(x) = 0. The value we seek will be a solution to ...
x = 3^(-1/4)
We can rearrange this to put it into an appropriate form.
x^4 = 3^-1 . . . . . . take 4th powers
3x^4 = 1 . . . . . . . .multiply by 3
3x^4 -1 = 0 . . . . . subtract the right side to make a polynomial in standard form
So, the function we want to find a zero of can be ...
f(x) = 3x^4 -1
The Newton's method iterator for this will be ...
[tex]x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\\\\=x_n-\dfrac{3x^4-1}{12x^3}=\dfrac{12x^4-3x^4+1}{12x^3}\\\\\boxed{x_{n+1}=\dfrac{9x^4+1}{12x^3}}[/tex]
In the attachment, we have defined the iteration function to be g(x). The graph shows us a starting value: [tex]x_0=0.76[/tex].
Then the first iteration will give accuracy to 3 decimal places*:
x ≈ (9·0.76^4 +1)/(12·0.76^3) = 4.00259584/5.267712 ≈ 0.7598357
The approximate result is x = 0.760.
_____
* The value shown here is actually accurate to 7 decimal places. One more iteration gets accuracy to the full precision of the calculator display.
