I was following these slides I found in Google since I couldn't understand my prof's lecture: https://www.math.ust.hk/~mamu/courses/231/Slides/ch02_2b.pdf
At slide 10, it shows five different transpositions of the function $f(x)=x^{3}+4x^{2}-10$. What I am interested in knowing is how did it get to $x = g(x) = x-\frac{x^{3}+4x^{2}-10}{3x^{2}+8x}$ since the root can be found within five iterations.
I assume that the denominator was determined from getting the first derivative of the numerator. Now, is this technique applicable to all cubic functions?
I also tried plugging values like '1' into the x variables to initiate a self-check in case I'm stuck. However, it gives out different values. For example, substitute 1 into 'x' in $\frac{1}{2}\sqrt{10-x^{3}}$ gives $\frac{3}{2}$, whereas using '1' to the equation I'm interested in gave a different answer, $\frac{16}{11}$.
The logic behind this is Newton's method. It does apply to all differentiable functions, not just cubics.
The basic idea is, given a function $f(x)$ and a particular point $x_n$ with $f(x_n)$ and $f'(x_n)$ specified, we can approximate the function using its Taylor series as $$ f(x) \approx f(x_n) + f'(x_n)(x - x_n) $$ If we try to find the root of the function on the RHS, assuming $f'(x_n) \ne 0$, you get $$ \begin{split} f(x_n) + f'(x_0)(x - x_n) &= 0 \\ x - x_n &= \frac{-f(x_n)}{f'(x_n)} \\ x &= x_n - \frac{f(x_n)}{f'(x_n)} \end{split} $$
Hence, we pick the next iteration $x_{n+1}$ as the RHS.