2 years ago

DON'T FORGET TO SIMPLIFY PLSSS!!!!!!

Mathematics

1 answer:

Bad White [126]2 years ago

5 0

Answer:

16692162

Step-by-step explanation:

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Given

$f(x)=x^3-x-1, x_1=1$

$f'(x)=3x^2-1$

Let the initial approximation is $x_1 =1$

So by Newton's method, we get

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},n=1,2,...$

$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{1^3-1-1}{3(1)^2-1}=1.5$

$x_3=1.5-\frac{(1.5)^3-1.5-1}{3(1.5)^2-1}=1.34782608$

$x_4=1.34782608-\frac{(1.34782608)^3-1.34782608-1}{3(1.34782608)^2-1}=1.32520039$

$x_5=1.32520039-\frac{(1.32520039)^3-1.32520039-1}{3(1.32520039)^2-1}=1.32471817$

$x_6=1.32471817-\frac{(1.32471817)^3-1.32471817-1}{3(1.32471817)^2-1}=1.32471795$

$x_5 \approx x_6$ are identical up to eight decimal places.

The approximate real root is x ≈ 1.32471795

∴ x = 1.32471795

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