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

  1. 2.222
  2. 2.275
  3. 2.279
  4. None of the above
in Numerical Methods by Boss (32.5k points)
edited by | 4.4k views

4 Answers

+5 votes
Best answer

Answer is (c) 2.279 

by Veteran (62.7k points)
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0
Is there any simple way to solve this type of questions?
0
how can solve this without calc??
+1

How We started from x0 = 3 ?

0
x0 is the initial root chosen so that it is closer to the required root..Since the options have positive integers, we can start with numbers 1,2,3 etc..Here,1 and 2 when substituted gives negative answers.And,.To get  xn+1,we use xn+1=xn-[f(xn)/f'(xn)]
+5 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) = x3-3x-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.

by Junior (515 points)
0
Your approach is easy but calculator is not allowed in ISRO exam..
0 votes
(c) 2.279
by
0 votes

c) by using this formula Xn+1 = Xn - (f(Xn) / f'(Xn))

by Boss (12.2k points)
+1
f(x)=x3 -3x-5

f(0)= -5<0

f(1)= -7<0

f(2)=  -3<0

f(3)=13 >0

so the real  root should lie between (2,3 )

why is that x0 = 2 cant be taken  ?

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