Now out of syllabus.

6 votes

In the Newton-Raphson method, an initial guess of $x_0= 2 $ is made and the sequence $x_0,x_1,x_2\:\dots$ is obtained for the function

$$0.75x^3-2x^2-2x+4=0$$

Consider the statements

- $x_3\:=\:0$
- The method converges to a solution in a finite number of iterations.

Which of the following is TRUE?

- Only I
- Only II
- Both I and II
- Neither I nor II

9 votes

Best answer

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ for Newton-Raphson method (See the link below)

$f(x) = 0.75 x^3 -2x^2-2x + 4$

$ \implies f(2) = -2$

$f'(x) = 2.25x^2 - 4x -2$

$\implies f'(2) = 9 - 8 - 2 = -1$

So, $x_1 = 2 - \frac{-2}{-1} =2-2 = 0$

$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0 - \frac{4}{-2} = 2$

Since $x_2 = x_0$, we will get $x_3 = x_1 = 0$.

So, $x_3 = 0$, and the method never converges. A choice.