I have an equation $l(x) = \sqrt{(x - 0.2)^2 + (x^2 - 2.7)^2}$.
Now I basically want to find at which x coordinate that $l(x)$ will be it's smallest.
I have differentiated the equation to find $l'(x) = (4x^3-8.8x-0.4)/\sqrt{(x - 0.2)^2 + (x^2 - 2.7)^2}$.
The next logical step I thought would be to find the roots of the derivative which will tell us the stationary points of the original equation. I tried Newtons Method for this as the question stated however I haven't been able to get anywhere near zero for it.
Can someone please state where I am going wrong here. A tutor told me that I have to find the derivative of the derivative to find the stationary points of the original function, however I can't make sense of that. I only thought that was relvant for classisfying the staionary points as a maximum, minimum or inflection point, which is after I find the staionary points?
(Sorry about the formatting, I did it all in word however it hasn't worked out. Could someone please direct me on how to insert symbols on this site?)
Since $\ell \ge 0$, you could work with $f(x) = l(x)^2$ instead, just to make algebra simpler. \begin{align} f(x) & = (x - 0.2)^2 + (x^2 - 2.7)^2 \\ f'(x) & = 2(x - 0.2) + 4x(x^2 - 2.7) \\ & = 2x - 0.4 + 4x^3 - 10.8x \\ & = 4x^3 - 8.8x - 0.4 \\ f''(x) & = 12x^2 - 8.8. \end{align} We see that $f''$ is not always positive. In fact, it will be positive only when $12x^2 - 8.8 > 0$, i.e., $x > \sqrt{\frac{8.8}{12}} = \sqrt{\frac{2.2}3}$ or $x < -\sqrt{\frac{2.2}3}$. Therefore, local minima are in the set $\left\{x \mid f'(x) = 0\right\} \cap \left(\left(-\infty, -\sqrt{\frac{2.2}3}\right) \cup \left(\sqrt{\frac{2.2}3}, \infty\right)\right)$. The problem now is to find $x$ such that $f'(x) = 0$, i.e., roots of $f'$.
The Newton method can be used to find roots of $f'$. The update equation is $$ x_{n+1} = x_n - \frac{f'(x_n)}{f''(x_n)} = x_n - \frac{4x_n^3 - 8.8x_n - 0.4}{12x_n^2 - 8.8}. $$ Since $f'$ is a polynomial of degree $3$, there are multiple roots. You should start iteration in the feasible set: $x_0 < -\sqrt{\frac{2.2}3}$ or $x_0 > \sqrt{\frac{2.2}3}$. This particular function $f'$ will give you two solutions, one from starting below $-\sqrt{\frac{2.2}3}$, and one from starting above $\sqrt{\frac{2.2}3}$. (I would pick $-2$ and $2$ as $x_0$ for simplicity.)
According to Wolfram Alpha, roots of $f'$ are approximately $-1.46$, $-0.045$ and $1.51$. $-0.045$ falls outside the feasible set (where $f''(x) > 0$), so the local minima are at $x \approx -1.46$ and $x \approx 1.51$. Computation would yield $f(1.51) < f(-1.46)$. (This agrees with Wolfram Alpha's optimization.)