Apply six decimal place rounding where applicableThe equation f(x) = x² - 6 = 0 has a root near x = 2.a) Apply four iterations of the Secant Method, taking the initial approximations x₁ = 1and x₂ = 2, to approximate the value of this root. Tabulate your iterations like inthe lecture notes.b) Determine the number of significant digits to which the last iteration's x3-valueapproximates the true solution of the equation.

Answered: The equation f(x) = x² - 6 = 0 has a…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Apply six decimal place rounding where applicable
The equation f(x) = x² - 6 = 0 has a root near x = 2.
a) Apply four iterations of the Secant Method, taking the initial approximations x₁ = 1
and x₂ = 2, to approximate the value of this root. Tabulate your iterations like in
the lecture notes.
b) Determine the number of significant digits to which the last iteration's x3-value
approximates the true solution of the equation.

Expert Solution

Step 1

Solution:-

$f (x) = x^{2} - 6 I s t I t e r a t i o n : - x_{0} = 1 a n d x_{1} = 2 f (x_{0}) = f (1) = - 5 a n d f (x_{1}) = f (2) = - 2 ∴ x_{2} = x_{0} - f (x_{0}) \cdot \frac{x_{1} - x_{0}}{f (x_{1}) - f (x_{0})} x_{2} = 1 - (- 5) \cdot \frac{2 - 1}{- 2 - (- 5)} x_{2} = 2.6667 ∴ f (x_{2}) = f (2.6667) = {2.6667}^{2} - 6 = 1.1111 2 n d i t e r a t i o n : x_{1} = 2 a n d x_{2} = 2.6667 f (x_{1}) = f (2) = - 2 a n d f (x_{2}) = f (2.6667) = 1.1111 ∴ x_{3} = x_{1} - f (x_{1}) \cdot \frac{x_{2} - x_{1}}{f (x_{2}) - f (x_{1})} x_{3} = 2 - (- 2) \cdot \frac{2.6667 - 2}{1.1111 - (- 2)} x_{3} = 2.4286 ∴ f (x_{3}) = f (2.4286) = {2.4286}^{2} - 6 = - 0.102$