Q2Advanced Engineering Mathematics I

Question

Find the approximate value correct to three places of decimal of the real root of the equation $x^{3} - 3 x + 4 = 0$ , using method of false position three times in succession.

Answer 1

Given $f (x) = x^{3} - 3 x + 4 = 0$ .

We first find an interval $[a, b]$ where $f (a)$ and $f (b)$ have opposite signs. Let $x = - 2 ⟹ f (- 2) = - 8 + 6 + 4 = 2$ . Let $x = - 3 ⟹ f (- 3) = - 27 + 9 + 4 = - 14$ .

So the root lies between $- 3$ and $- 2$ .

The Regula Falsi formula for the next approximation is: $x_{n + 1} = \frac{a \cdot f ( b ) - b \cdot f ( a )}{f ( b ) - f ( a )}$

Applying this three times in succession yields the root correct to three decimal places. Each iteration replaces either $a$ or $b$ depending on the sign of $f (x_{n + 1})$ .

Q2Advanced Engineering Mathematics I

Question

Answer