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?
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!
Hint: I have got this here
$$f'(x)={\frac {6\,{x}^{3}-16\,{x}^{2}+13\,x-26}{\sqrt {4\,{x}^{2}+13}}}$$