Numerical method to find the roots of the derivative of $y = (x^3 - 4x) \cdot \frac{\sqrt{13+4x^2}}{2}$ (Newton's Method) I am working on an Optimization problem and I would like to suggest finding the solution (i.e. the extrema) of the function below using two methods: the first one being calculating the derivative and making it equal to zero and the second one using a numerical method.
The equation of the model is shown below:
\begin{align*}
y = (x^3 - 4x) \cdot \frac{\sqrt{13+4x^2}}{2} 
\end{align*}
(the derivative being the following)
\begin{align*}
y' = \frac{16x^4 +7x^2 - 52}{2 \sqrt{13+4x^2}}
\end{align*}
I have two questions about using it:


*

*Based on some research, I found it is possible to use Newton's Method on functions that are not polynomials. Is that correct?

*Is the Newton's Method the most adequate in this case? 

 A: We want to find the extrema of the function
$$\begin{align*}
y(x) = (x^3 - 4x) \cdot \frac{\sqrt{13+4x^2}}{2} 
\end{align*}$$
One approach is to use Newton's Method on $y'(x)$, where
$$y'(x) = \frac{16 x^4+7 x^2-52}{2 \sqrt{4 x^2+13}}$$
We start by plotting $y(x)$ and see two extrema to try and find using Newton's 
The iteration formula for Newton's Method is given by
$$x_{n+1} = x_n-\dfrac {f(x_n)}{f'(x_n)} = x_n - \dfrac{16x_n^4  + 7x_n^2 - 52}{2 \sqrt{4 x_n^2 + 13}\left( \dfrac{64 x_n^3 + 14 x_n}{2 \sqrt{4 x_n^2 +13}} - \dfrac{2 x_n (16 x_n^4 + 7 x_n^2 - 52)}{(4 x_n^2 + 13)^{3/2}}\right) }$$
This can be simplified to remove the square roots and nasty divisions as
$$x_{n+1} = x_n - \dfrac{\left(4 x^2+13\right) \left(16 x^4+7 x^2-52\right)}{2 x \left(96 x^4+430 x^2+195\right)}$$
If we initialize Newton's method using $x_0 = -1.4$, we arrive at $x_4 = -1.26382$ in $4$ steps.
If we initialize Newton's method using $x_0 = 1.4$, we arrive at $x_4 = 1.26382$ in $4$ steps.
Update
There is an easier approach to find where $y'(x) = 0$, we need only find the roots of the numerator of $y'(x)$, so the Newton iteration can be simplified to
$$x_{n + 1} = x_n - \dfrac{16 x_n^4 + 7 x_n^2 - 52}{64 x_n^3 + 14 x_n}$$
This gives the same results as before. Also, comparing this with the above iteration, the quadratic term $(4 x^2 + 13)$, leads to imaginary roots that we don't care about. 
Lastly, it is worth noting that you can use a simple transform, $t = x^2$, on the iterations' numerator to get $16 t^2 + 7 t - 52 = 0$, and then solve a quadratic formula and eliminate Newton's Method altogether!
A: Hint: I have got this here
$$f'(x)={\frac {6\,{x}^{3}-16\,{x}^{2}+13\,x-26}{\sqrt {4\,{x}^{2}+13}}}$$
