Given that the function has a minimum value of 2 at , determine the values of and .
First, find .
Since is defined for all , the critical points of occur when .
It is given that has a minimum value of 2 at , so when , it must be true that .
For this reason, substitute 0 for and for in the equation and solve for .
Now substitute 2 for , for , and 2 for in the equation , then solve for .
Thus, the value of is 2, and the value of is 3.