Find the maximum and minimum values, for x ∈ [0, 3], of:
f (x) = 3x² − 6x + 5
The function is differentiable everywhere, and f^{\prime}(x) = 6x − 6 = 6(x − 1). Hence x = 1 is the only critical point. The candidate extreme points are the end points 0 and 3, as well as x = 1. We calculate the value of f at these three points. The results are f (0) = 5, f (3) = 14, and f (1) = 2. We conclude that the maximum value is 14, obtained at x = 3, and the minimum value is 2 at x = 1.