6 votes

Using Newton-Raphson method, a root correct to 3 decimal places of $x^3 - 3x -5 = 0$

- 2.222
- 2.275
- 2.279
- None of the above

5 votes

8 votes

**Shortcut to find an ****answer** to such question in an exam.

**Newton-Raphson method is** a method for finding successively better approximations to the roots (or zeroes) of a real-valued function. It is one example of a **root-finding algorithm.**

Here f(x) = x^{3}-3x-5

Just put the options one by one in f(x) and check whether we are getting "0.000..."(correct up to 3 decimal places) as the answer.

op A: x= 2.222 ie. f(x) = (2.222)^{3}-3(2.222)-5 = -0.695354... // op A is not correct

op B: x= 2.275 ie. f(x) = (2.275)^{3}-3(2.275)-5 = -0.050453... // op B is not correct

op C: x= 2.279 ie. f(x) = (2.279)^{3}-3(2.279)-5 = -**(0.000)**236361... // op C is correct root of given equation (upto 3 decimal places..

So** Op C is the answer.**