Input Resistance vs Output Resistance I am having trouble understanding the difference between these two concepts. I know that input resistance is the resistance as seen by the input terminals and output is as seen at the output terminals. What I don't understand is how these could end up being different values. I know how to find the Thevenin resistance as viewed from either terminal, but as far as I know this is only the correct approach for the output.

For example, in the image above, why is the input resistance not simply Rs and the output resistance Rl?
 A: 
I know that input resistance is the resistance as seen by the input
  terminals

As written, this isn't correct.  The input resistance is the resistance looking into the input terminals.  Conceptually, this means that if one changes the voltage across the input terminals (only), the input current changes by
$$\Delta i_{i} = \frac{\Delta v_{i}}{R_{i}}$$
Similarly, if one changes the voltage across the output terminals (only), the output current changes by
$$\Delta i_{o} = \frac{\Delta v_{o}}{R_{o}}$$
This essentially defines the input and output resistance.  Of course, there are different but equivalent ways to define and measure these parameters.
It easy to see how they could be different values.  For example, the input resistance can be infinite (open circuit) while the output resistance can be zero.  This is the case for an ideal voltage amplifier:

Clearly, there is no path for input current so the input current is zero which implies infinite input resistance.  The output terminals connect to an ideal (controlled) voltage source which can supply any current and thus, the output resistance is zero.
A: Input resistance and output resistance are parameters to help you use a particular circuit.
In your example above the attenuator starts at Port 1 and finishes at Port 2.
Removing the load resistor $R_L$ for the moment suppose you were asked what the input voltage to the attenuator $V_i$ was with the given supply.
To answer such question you could reduce all those resistors between Port 1 and Port 2 to one resistor $R_i$.  This resistor is called the input resistance of the circuit and if you knew that at the start your evaluation of $V_i$ would have been comparatively easy in that the whole of the circuit (ignoring $R_L$) would be reduced to a voltage source $V_S$ in series with two resistors $R_S$ and $R_i$ which form a potential divider network.
$V_i =\dfrac {R_i}{R_i+R_S}\cdot V_S$
This is where @Alfred Centauri's method of measuring input resistance comes from.  $V_S$ is equally divided between $R_S$ and $R_i$ when $R_S = R_i$.
In the example which is given with $R_L$ also present then the input resistance would change $R^\prime_i$.  So the input resistance of a circuit sometimes depends on what is connected across the output terminals.  
Active amplifiers often have a number of input each of which might have a different input resistance (or impedance) so as the source can be matched to the amplifier so that the signal from the external source is input into the amplifier in the most efficient way.
The output resistance can indeed be thought of as a Thevenin equivalent resistance and again in this example the output resistance of the attenuator between Port 1 and Port 2 will differ if the source is added across the input terminals.
Knowing $R_o$ would make it easier to compute what fraction of the voltage is to be found across $R_L$.
For active amplifier one would get maximum power tanswer if $R_o$ was equal to the resistance (impedance) of the loudspeakers connected to the output.
