Time period of bar pendulum

Why does the time period of a bar pendulum first decrease and then increase when the distance of the axis of rotation from the centre of gravity increases?

The period of a pendulum executing small angle oscillations is given by $$T = 2\pi\sqrt{\frac{\mathcal{I}}{mgr}}$$ where $\mathcal{I}$ is the moment of inertia about the axis of rotation and $r$ is the distance from the center of mass to the axis of rotation.
For your situation, use $r$ as a variable. The moment of inertia can be found using the parallel axis theorem to be $$\mathcal{I}=m\left( \frac{\ell^2}{12}+r^2\right)$$, where $\ell$ is the length of the bar.
Plot $T$ as a function of $r$ and you'll have your answer. Arbitrarily choose a value for $\ell$ (1 or 2). Start $r$ at some small value. Mathematically, $\mathcal{I}$ is increasing in the numerator while $r$ is increasing at a different rate in the denominator.