Question

Find the root of equation x³-x-4 = 0, correct to three decimal places using Newton Raphson method.​

Answer 1

Answer:

The root of equation x³-x-4 = 0 to three decimal places using the Newton Raphson Method is 1.797

Step-by-step explanation:

According to the Newton Raphson Method:

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

where $x_{n}$ is starting guess.

$f(x) = x^{3}- x- 4$

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

Starting guess is $(x_{o} = 2)$

Now first iterations,

$x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} \\\\\\x_{1} = x_{o} - \frac{[{x_{o}]^{3}} - x_{o} - 4}{3[x_{o}]^2 - 1}\\\\x_{1} = 2 - \frac{2^{3} -2 - 4}{3(2^{2}) - 1} = 2 - \frac{2}{11} = 1.818$

Now second iterations,

$x_{n+2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})} \\\\\\x_{2} = x_{1} - \frac{[{x_{1}]^{3}} - x_{1} - 4}{3[x_{1}]^2 - 1}\\\\x_{2} = 1.818 - \frac{1.818^{3} -1.818 - 4}{3(1.818^{2}) - 1} = 1.818 - \frac{0.191}{8.916} = 1.797$

Therefore,

$x_{2} = 1.797$

Answer 2

Answer:

1.796

Step-by-step explanation:

Here x3-x-4=0

Let f(x)=x3-x-4

d/dx(x3-x-4)=3x^2-1

∴f′(x)=3x^2-1

Here

x     0    1  2 

f(x) -4  -4  2

Here f(1)=-4<0andf(2)=2>0

∴ Root lies between 1 and 2

x0=1+2/2=1.5

x0=1.5

1st iteration :

f(x0)=f(1.5)=1.5^3-1.5-4=-2.125

f′(x0)=f′(1.5)=3⋅1.5^2-1=5.75

x^1=x0-(f(x0)/f′(x0))

x^1=1.5-(-2.125/5.75)

x^1=1.8696

2nd iteration :

f(x1)=f(1.8696)=1.8696^3-1.8696-4=0.6651

f′(x1)=f′(1.8696)=3⋅1.8696^2-1=9.4858

x^2=x1-f(x1)/f′(x1)

x^2=1.8696-(0.6651/9.4858)

x^2=1.7995

3rd iteration :

f(x^2)=f(1.7995)=1.7995^3-1.7995-4=0.0272

f′(x^2)=f′(1.7995)=3⋅1.7995^2-1=8.7141

 x^3=x^2-(f(x2)/f′(x2))

x^3=1.7995-(0.0272/8.7141)

x^3=1.7963

4th iteration :

f(x^3)=f(1.7963)=1.7963^3-1.7963-4=0.0001

f′(x^3)=f′(1.7963)=3⋅1.7963^2-1=8.6804

x^4=x3-(f(x3)/f′(x3))

x^4=1.7963-(0.0001/8.6804)

x^4=1.7963

Approximate root of the equation x3-x-4=0 using Newton Raphson mehtod is 1.7963 (After 4 iterations)