Questions about the baxandall circuit
Your circuit is indeed a tone control circuit. 



For low signal frequency (Bass booster) we can ignore \$C_2\$ and \$C_1\$ because for low-frequency capacitance reactance \$X_C\$ is high. 

So we have this situation (max bass boost): 



And bass boost gain is: 

$$A_{Vmax} = \frac{R_2+P_1}{R_1}$$

And as signal frequency increases \$C_2\$ capacitor reactance decreses.
And for the frequency when \$X_{C2} = P1\$ the bass boost gain starts to decrees.
Assuming \$R_1 = R_2\$

\$ \Large F_{LP} \approx \frac{0.16}{P_1 \cdot C_2}\$ 

And the bass gain will reach \$1V/V (0dB)\$ around this frequency

\$ \Large F_{LZ} \approx \frac{0.16}{R_1 \cdot C_2}\$       

The situation for high frequency (treble booster) looks slightly more complicated. 

Additional assuming that the bass pot is "set" at half (\$0.5P_1\$) we have this situation:



And to see what is going on you can replace \$C_1\$ capacitor with a short circuit. 
and if you do that you will see almost classic version of an inverting amplifier.   

Anditional if we choose the resistor so that \$R_4 &lt;&lt; (R_1 + 2R_S + 0.5P_1)\$

We have:

\$ \Large F_{HP} \approx \frac{0.16}{(R_1 + 2R_S + 0.5P_1) \cdot C_1}\$  

And 

\$ \Large F_{HZ} \approx \frac{0.16}{R_4\cdot C_1}\$  

And the treble boost max gain is: 

$$A_V \approx \frac{R_1 + 2R_S + 0.5P_1}{R_4}$$

And of course \$R_X = R_1 + 2R_S + 0.5P_1\$