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

f'(x) = 6x^{2} - 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) = 2x^{3} -3x^{2} - 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.