I want to find the maximum value of the following function \begin{align} f(t)=e^{-\frac{t^2}{2}}\frac{| e^{-ta}(1+at)+e^{ta}(1-at)|}{\sqrt{e^{-2ta}+e^{2ta}}} \end{align} for some $a>0$.
This function is symmetric around zero and has three local maximum ( from the graph) one of which is at zero. See the figure below.
However, the lobs change their values, depending on what $a$ is, and the maximum doesn't alway happen at $x=0$.
I also try finding derivative at $f(t)$ but it is way to difficult to handle it, see here.
I propose that $\eta = at$ to make things appear more clear. Secondly use: $$\cosh{x}=\frac{1}{2}[\exp{(x)}+\exp{(-x})] \qquad \sinh{x}=\frac{1}{2}[\exp{(x)}-\exp{(-x})]$$ Then: $$f(\eta) = \sqrt{2}\,\exp{\left(-\frac{\eta^2}{2a^2}\right)}\frac{\cosh{(\eta)}-\eta\,\sinh{(\eta)}}{\sqrt{\cosh{(2\eta)}}}\tag{*}$$
I get rid of the absolute value because the all the extrema of $(*)$ are maxima of your original function. Since the function $f(\eta)$ is symmetrical wrt. the vertical axis you can look for only the extrema for $\eta\geq 0$, and what is more $f(\eta)$ cuts once the $\eta$ axis approximating to $0$ from negative values. Therefore the function will have for $\eta\geq 0$ two extrema: one maximum at $\eta=0$ (because $\cosh{(\eta)}-\eta\sinh{(\eta)}$ has a maximum for $\eta=0$) and one minimum which will move further from $0$ as $a$ does.
The extrema of $(*)$ cannot be found analytically (I guess), therefore I recommend you to use some numerical algorithm like Newton-Raphson to find the roots for a particular $a$.
$$\eta^{n+1}=\eta^n-\frac{f'(\eta^n)}{f''(\eta^n)}$$
It is easy, but tedious to calculate the derivatives of $(*)$. Since you are looking for one minimum only, you can begin with some arbitrary large enough seed for $\eta$. This value will tend always to the correct solution. Try for example $\eta=\sqrt{2}a$.