Suppose f(x) = x^2 - 3x + 7, which is a parabola, and want to know the maximum and minimum on the interval [0,2].
First, we find the critical values, which is where the slope along the curve equals 0. To get that we take the derivative, set equal to 0 and solve for x.
f'(x) (derivative) = 2x - 3.
Set equal to 0 and we see that x = 3/2 or 1.5
Now, find the y coordinate of the endpoints of the interval and of the critical point.
f(0) = 7
f(2) = 5
f(1.5) = 4.75
Therefore the maximum on the interval is at 0,7 and the minimum on the interval is (1.5, 4.75). In fact, since this is a parabola, the vertex is (1.5, 4.75), which is the minimum value of the curve.