this is the kind of stuff I was trying to explain qualitatively to a fellow ee classmate several quarters back.
Voltage cannot change instantly in a capacitor and current cannot change instantly in an inductor.
So for this simple circuit, the capacitor with zero initial charge (0V) will have 0 volts at t = 0- and 0 volts at t = 0+. It charges up to V through the resistor R exponentially with a time constant of R*C. At t = x-, when the ideal capacitor is charged to V volts, it will remain at V volts at t = x+. Therefore the voltage across the resistor at t = x+ will be V - Vcap, or -V. The capacitor will then discharge through R exponentially with a time constant of R*C.
Replace the capacitor with an inductor and you can do a similar analysis. Since the current cannot change in the inductor instantly, and it is in series with the resisitor, then the voltage across the resistor at t = 0- is the same as that at t=0+, or 0 volts. In fact if you look at the voltage across the inductor, it will look similar to the voltage across the resistor in the capacitor example (for certain values of L and R). The voltage across R will look similar to the voltage across the capacitor in the capacitor example also. This duality (is that the correct word?) is pretty dang cool.
Btw, I failed at trying to help my fellow student realize this, and they just went back to trying to memorize formula to work all of the different circuit topologies that might be on the exam. It turned out the exam was simple enough that this worked.