Question

Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your answer to four decimal places.) 2 x − x2 + 1 = 0, x1 = 2

Answer 1

Answer:

$x_{3}=2.35$

Explanation:

Given $x^2-2x-1=0,x_1=2$

From Newton's method

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

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

$f'(x)=2x-2$

Now

$x_{2}=x_1-\dfrac{f(x_1)}{f'(x_1)}$

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

$f(2)=2^2-2\times 2-1$

f(x)=-1

$f'(2)=2x-2$

$f'(1)=2\times 2-2$

f'(1)=2

$x_{2}=2+\dfrac{1}{2}$

$x_{2}=2.5$

$x_{3}=x_2-\dfrac{f(x_2)}{f'(x_2)}$

$f(2.5)=2.5^2-2\times 2.5-1$

f(2.5)=0.45

$f'(2.5)=2\times 2.5-2$

$f'(2.5)=3$

$x_{3}=2.5-\dfrac{0.45}{3}$

So

$x_{3}=2.35$