Solution We have
f (x) = x3 – 3x + 3
or
f ′(x) = 3x2 – 3 = 3(x – 1) (x + 1)
or
f ′(x) = 0 at x = 1 and x = – 1
Thus, x = ± 1 are the only critical points which could possibly be the points of local
maxima and/or local minima of f Let us first examine the point x = 1 Note that for values close to 1 and to the right of 1, f ′(x) > 0 and  for values close
to 1 and to the left of 1, f ′(x) < 0 Therefore, by first derivative test, x  = 1 is a point
of local minima and local minimum value is f (1) = 1