Maximizing $e^{-x^2/2}\cosh(ax)$

Question

Consider the function $f_a:\mathbb R \to \mathbb R$ given by $$ f_a(x) = e^{-x^2/2}\cosh(ax). $$ It seems that there is a value of $c > 0$ such that for $a \in (0,c)$ the maximum of the function occurs at $0$ and for $a \in (c,\infty)$ the maximum occurs and values $\pm M_a$ where $M_a$ is nonzero. What is that threshold value $c$ (if the above is true)?

Answer 1

\begin{align} f_a'(x) & = e^{-x^2/2}(-x)\cosh(ax) + a e^{-x^2/2} \sinh(ax) = 0 \\ & \implies a\tanh(ax) = x. \end{align} We may as well consider $a > 0$. Clearly $x = 0$ is always a critical point and note that the LHS approaches $\pm 1$ as $ x \to \pm \infty$. I will leave it to you to show that if $0 < a < 1$ then the above equation only has the solution $x = 0$.