Given parameters:
- L - the lowest point,
- R - the point where chart crosses zero,
- b - some coefficient;
a function $f$ is defined as follows:
$$f(x) = ( \frac {L} {R^6} x^6 - L) e ^ {-bx^2}$$
A chart for this function ($L=0.5$, $R=1$, $b=1$), looks like this:
How to find the $x$ of a maximum point, where derivative of $f$ crosses zero?
I can see that solution is at $x~=1.7$, but I failed to solve it analytically...
Do you know, how to solve it? Or is analytical solution possible at all?
Compute the derivative of $f(x)$ and set it to zero,
$$f'(x) = e ^ {-bx^2} \frac {L} {R^2} \left[-2b(x^6-R^2)x+6x^5\right] =0$$
or, with $R=1$ and $b=1$,
$$x\left(x^6-3x^4-1\right)=0$$
which has the obvious solution at $x=0$ for the minimum value. Note that the cubic equation in $x^2$ has only one real root and it is given analytically by
$$x^2= 1+\left(\frac{3-\sqrt5}{2}\right)^{1/3} +\left(\frac{3+\sqrt5}{2}\right)^{1/3}$$
Thus, the maximum values are at $x =\pm 1.762$.