Answer to Question #308200 in Quantitative Methods for raf

Question

find the root between (2,3) of x^3-2x-5=0, by using false position method

Answer 1

According to the false position method

$x_n = x_1 - \frac{x_{n-1}-x_1}{f(x_{n-1}-f(x_1)}f(x_1)$

In our case $f(x) = x^3-2 x-5$ , and x₁ = 2, x₂ = 3

$\begin{matrix} n & x_n &f(x_n) \\ 1& 2& -1 \\ 2& 3& 16 \\ 3& 2.058823529& -0.390799919 \\ 4& 2.096558637& 0.022428062 \\ 5& 2.094440519& -0.001238424 \\ 6& 2.094557621& 6.85303\cdot10{-5} \\ 7& 2.094551142& -3.79179\cdot10^{-6} \\ 8& 2.0945515& 2.09802\cdot 10^{-7} \\ \end{matrix}$