Is inertia and gravity determined by relativistic mass or invariant mass? As far as I know, mass fundamentally determines inertia and the gravitational force. But since there are two types of mass, which mass determines which? From what I have read so far, and correct me if I'm wrong, the relativistic mass determines the inertia, but not the gravitational force. Then why does one determine inertia and another determine gravity? Also since relativistic mass represents the total mass-energy of an object taking into account the kinetic energy, does that imply that the gravity is not determined by the total energy content of an object, but only by its invariant mass which doesn't take into account its kinetic energy?
 A: Neither inertia nor gravity is determined by either the mass $m$ or by the so-called relativistic mass $m\gamma$.
If you write Newton's second law in terms of the acceleration three-vector and the force three-vector, it looks like $F=m\gamma a_\perp+m\gamma^3 a_\parallel$. Although it's true that you can write the second law in terms of four vectors as $F=ma$, the four-vector force is not the force that any observer actually measures, and it doesn't behave the way newtonian forces behave for purposes of computing work (its inner product with the velocity is always zero).
The source of gravity is the stress-energy tensor, not a scalar such as $m$ or a single real number such as $m\gamma$.
BTW, relativistic mass is becoming deservedly extinct. It's no longer used in writing by professional physicists or in textbooks. It's only used these days in popularizations. See Oas, "On the Abuse and Use of Relativistic Mass," 2005, http://arxiv.org/abs/physics/0504110 .
A: As stated is not advised to call the relativistic energy $m\gamma$ relativistic mass. Mass these days strictly refers to total energy in the rest frame divided by $c^2$.
$m\gamma$ is indeed the source of gravity and it determines inertia, as the relativistic momentum is $m\gamma \vec v$. Newton's second law is replaced by $$\dot {\vec p} = m \gamma {\vec a} + \gamma^3 \left( {\vec v} \cdot {\vec a} \right) {\vec v} ~.$$
