The questions reads as follows:
Solve for the root using Fixed Point Method with initial guess of $T_{f} = -100, \varepsilon_{s}=0.001%$ $f(T_{f})=-0.50598\times 10^{-10}T_{f}^{3}+0.38292\times 10^{-7}T_{f}^{2}+0.74363\times 10^{-4}T_{f}+0.88318\times 10^{-2}$
I understand the process of this method, and I know that I have to test the derivative of the given function first to $\left | g'(x) \right |<1$. I also understand that the given $T_{f}$ is the initial guess.
What I am worried about is finding the $T_{f}$. I tried several algebraic techniques to get it, including $T_{f} = \frac{-0.38292\times 10^{-7}T_{f}^{2}-0.74363\times 10^{-4}T_{f}-0.88318\times 10^{-2}}{-0.50598\times10^{-10}}$. However, this, among many attempts, after differentiating it, failed to meet $\left | g'(x) \right |<1$ to say that it converges.
What would be the best approach to solve this dilemma?
You get good results by solving for the highest power and then taking the corresponding root on both sides, (using $x=10^{-3}T_f$ to use the obvious scaling of the coefficients) $$ x_{n+1}=-\sqrt[3]{-\frac{0.38292x_n^2+0.74363x_n+0.88318×10^{−1}}{0.50598}}. $$ For $|x_n|<1$ the linear term dominates the other two so that the sign of $x_n$ determines the sign of the quotient under the root.
But actually, you can use the Cardano formulas in the numerical version, that is using trigonometric or hyperbolic angle trisectors. Or use the Newton method. You could probably even use your iteration formula as input for the Aitken delta-squared process. A direct polynomial solver like the Jenkins-Traub method will also find the roots sufficiently fast. They are $$ -128.75486193404798, -802.9052672568066, 1688.44893507251 $$