Before proceeding, note that this is an upwards facing parabola, meaning we can know without further calculation that it will have no maxima, and a single minimum at its vertex. Completing the square would show us that f(x) = 3(x-2)^2+1, giving the vertex, and thus the sole minimum, at x = 2. Let's see how this would be done with calculus, though.
Any extrema will occur either at a critical point or at an endpoint of the given interval. As our given interval of (-oo,oo) is open, we can ignore the possibility of endpoints, and so we will first identify the critical points of the function, that is, the point at which the derivative of the function is 0 or does not exist.
f'(x) = d/dx (3x^2-12x+13) =6x-12
Setting this equal to 0, we find a critical point at x=2
6x-12 = 0 => x = 12/6 = 2
Now, we can either test to see whether it is an extremum (and what type) by checking some values of f around that point, or by using the second derivative test. Let's use the latter.
(d^2x)/(dx^2) = d/dx(6x-12) =6
As f''(2) = 6 > 0, the second derivative test tells us that f(x) has a local minimum at x=2
Thus, using f'(x) and f''(x), we find that f(x) has a minimum at x=2, matching the result we found using algebra.
