Problem : f (x) = 2x3 -3x2 - 4. Use the second derivative test to classify the critical points.

f'(x) = 6x2 - 6x;
f'(x) = 0 at x = 0 and x = 1.
f''(x) = 12x - 6;
f''(0) = - 6, so there is a local max at x = 0.
f''(1) = 6, so there is a local min at x = 1.

Problem : Describe the concavity of f (x) = 2x3 -3x2 - 4 and find any inflection points.

f''(x) = 12x - 6, so f''(x) = 0 when x = . The sign of f''(x) and the concavity of f are depicted below: f has an inflection point at x = because the concavity of the graph changes there.

Problem : f (x) = sin(x). Use the second derivative test to classify the critical points on the interval [0, 2Π].

f'(x) = - cos(x);
f'(x) = 0 at x = and x = .
f''(x) = - sin(x);
f''( ) = - 1, so f has a local maximum there.
f''( ) = 1, so f has a local minimum there.

Problem : Describe the concavity of f and find any inflection point for f (x) = sin(x) on the interval [0, 2Π].

f''(x) = - sin(x), so f''(x) = 0 at x = 0, x = Π, and x = 2Π. The sign of f''(x) and the concavity of f are depicted below: On the interval [0, 2Π], f only has an inflection point at x = Π. If we were to extend the graph in both directions, x = 0 and x = 2Π would also be points of inflection.