$f(x) = e^{x} f(-x)$ with f(x) > 0

Is there anything I can say about the general shape of this function (defined on the real axis)? For example the formula gives the derivative of f in zero in terms of f(0) (which is okay assuming I'm only interested in f up to a multiplicative constant).
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 EDIT: Note by the way that a non-trivial solution is $f(x) = e^{x/2}$ (although I'm more interested in normalizable solutions) EDIT2: Another non-trivial solution is $f(x) = e^{-(x-1)^2/4}$
 Alright I think I've found the general solution. Since f(x) > 0, we can write $f(x) = e^{g(x)}$. The functional equation becomes $e^{g(x) - g(-x)} = e^x$ such that $g(x) = g(-x) + x$. If we assume we can write g(x) as a power series, we have $g(x) = \sum a_n x^n$. Substituting it in the functional equation: $\left\{ \begin{array}{ll} a_1 = \frac{1}{2} & \\ a_{n} = 0 & \textrm{for n > 1 odd} \end{array} \right.$ The coefficients for the even powers are arbitrary. If we want the solution to be normalizable, we only need to demand that the coefficient of highest power is even with negative coefficient. More general: $f(x) \propto \exp \left( \frac{x}{2} + \sum_{n=1}^{+ \infty} a_{2n} x^{2n} \right)$ EDIT: perhaps a more insightful formulation $f(x) = e^{x/2} \cdot h(x)$ with h > 0 an even function.

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 Quote by mr. vodka $g(x) = g(-x) + x$. If we assume we can write g(x) as a power series, …