Are you referring to a specific material or a physical phenomenon? As such, the dielectric constant decreases with increasing frequency, as the net polarization in the material drops (because the dipoles can't "keep up" with the fast changing fields). The dielectric constant will also generally drop as the temperature goes up, again due to the dipoles "losing orientation" due to large thermal noise.
Your question is kind of loaded actually, because you're trying to vary both temperature and frequency.
Here's a helpful animation for dielectric constant vs frequency: http://www.doitpoms.ac.uk/tlplib/dielectrics/variation.php
And temperature: http://www.doitpoms.ac.uk/tlplib/dielectrics/temperature.php
At high frequency the heavy dioples (e.g. ionic polarizability) canot follow the high frequency field variation and stop participating in the total polarization (ionic and electroinc polarizabilities), thus at high frequency only the electronic polarizability is the effective one and the dielectric constant (relative permittivity) value will drop.
At high temperature another effect will take place, heat will scatter the dipoles to some extent depends how high the temperature is, and will cause an additional drop in the relative permittivity.
I suggest you do two experiments and measure the relative permittivity:
1. Keep the temperature constant and vary the frquency of the applied source, and see the variation of the rel. permittivity.
2. Keep the the frequency constant and vary the temperature and calculate the rel. pemittivity.
Discuss the results and get the proper conclusion.
You need to specify the meaning of "high frequency" as well as the material type, e.g., in the region of bandgap or optical phonon frequency? is this an ionic material?