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.