Rolle’s theorem says that 1) if a function is continuous an interval (in this case [0, 3] and 2) differentiable on the open interval (in this case (0, 3)) and if f(0) = f(3), then there will at least one value of x such that f ′(x) = 0 on the interval.
So, let's check the conditions. 1) Yes, x2 - 3x + 1 is both continuous and 2) differentiable (it is so any real number). f(0) = 1 and f(3) = 1 so all conditions have been met.
Remember that x2 - 3x + 1 is a parabola. At the vertex, the derivative will equal zero. The vertex is located at x = -b/2a (from algebra) or x = 3/2 = 1.5 and x = 1.5 is located on the interval. Since the vertex is the ONLY place on a parabola where the derivative is zero, we are finished. c = 1.5