Question

Use the Newton’s method to approximate the real solution of xcube-2x-2=0 in the interval [1,2]

Answer 1

x³ - 2x - 2 = 0 x $\in$ [ 1, 2 ]

Let f(x) = x³ - 2x - 2

f(1) = (1)³ - 2(1) - 2

f(1) = -3

f(2) = (2)³ - 2(2) - 2

f(2) = 2

Hence, a real solution of f(x) exists between 1 and 2.

f ' (x) = 3x² - 2

Let x₀ = 1.5

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

$\implies$ n= 0

x₁ = x₀ - $\dfrac{f(x\scriptscriptstyle 0)}{f'(x\scriptscriptstyle 0)}$

x₁ = 1.5 - $\dfrac{f(1.5)}{f'(1.5)}$

f(1.5) = (1.5)³ - 2(1.5) - 2 = -1.625

f ' (x) = 3(1.5)² - 2 = 4.75

x₁ = 1.5 + $\dfrac{1.625}{4.75}$

x₁ = 1.842

$\implies$ n =1

x₂ = x₁ - $\dfrac{f(x\scriptscriptstyle 1)}{f'(x\scriptscriptstyle 1)}$

x₂ = 1.842 - $\dfrac{f(1.842)}{f'(1.842)}$

f(1.842) = (1.842)³ - 2(1.842) - 2 = 0.566

f ' (x) = 3(1.842)² - 2 = 8.178

x₂ = 1.842 - $\dfrac{0.566}{8.178}$

x₂ = 1.772

$\implies$ n =2

x₃ = x₂ - $\dfrac{f(x\scriptscriptstyle 2)}{f'(x\scriptscriptstyle 2)}$

x₃ = 1.772 - $\dfrac{f(1.772)}{f'(1.772)}$

f(1.772) = (1.772)³ - 2(1.772) - 2 = 0.0201

f ' (x) = 3(1.772)² - 2 = 7.420

x₃ = 1.772 - $\dfrac{0.0201}{7.420}$

x₃ = 1.769

$\implies$ n = 3

x₄ = x₃ - $\dfrac{f(x\scriptscriptstyle 3)}{f'(x\scriptscriptstyle 3)}$

x₄ = 1.769 - $\dfrac{f(1.769)}{f'(1.769)}$

f(1.769) = (1.769)³ - 2(1.769) - 2 = -0.00216

f ' (x) = 3(1.769)² - 2 = 7.388

x₄ = 1.769 - $\dfrac{-0.00216}{7.388}$

x₄ = 1.769

Now we see that x₄ = x₃ . Hence, the root of the given equation is 1.769 correct up to three decimal places.