Problem 8.6-4<

The only thing in this problem is to understand that before t = 0, the voltage is 6 + 6 = 12 V, and after t = 0, it is 6 V

We then have a voltage divider that we can apply here to find the voltage Vc(t).

Before t = 0 we have Vc(t) = 12*4/(2+4) = 8 V

After t = 0, we get Vc(t) = 6*4/(2+4) = 4 V