A probability density function with sharp turn at its mode $(f''(\text{mode})=-\infty)$ Does there exist a known (with name) probability density function $f$ with mode $x_0$ where $f''(x_0)=-\infty$ and also one has

*

*$f$ and $f'$ are both continuous (and hence $f'(x_0)$ exists and is equal to zero)

*$f$ and $f''$ are symmetric around $x_0$
Preference is for a probability distribution with bounded support.
Update: To be more specific I'm trying to find a PDF which matches below graph (the graph is derived from a very complicated ODE and hence I should "guess" the solution). The ones with absolute values over 10 for mode are basically infinity.

 A: There is the Laplace distribution.
Sometimes people call this the "double exponential distribution", but that term is also used for a distribution whose density is an iterated exponential distribution. I think I've also seen this called a bilateral exponential distribution.
The function $f,$ rather than being called a probability distribution, should be called a probability density function. But the expression $f(x) \, dx$ (with $\text{“}dx\text{''}$ may be called a probability distribution since it can be identified with the mapping
$$
A\mapsto \int_A f(x)\,dx
$$
for (Borel) sets $A.$ Thus you could say:

Does there exist a known (with name) continuous  symmetric (around its mode $x_0$) probability distribution whose density function $f$ satisfies $f''(x_0)=\infty$?

If you want a bounded support, there is the triangular distribution
$$
f(x)\,dx= (1 - |x|)\,dx \quad \text{for } 0<x<2.
$$
This is the distribution of the sum of two independent random variables that are uniformly distributed in the interval $(0,1).$
