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Problems for "Using the Second Derivative to Analyze Functions"

Problems for "Using the Second Derivative to Analyze Functions"

Problems for "Using the Second Derivative to Analyze Functions"

Problems for "Using the Second Derivative to Analyze Functions"

Problems for "Using the Second Derivative to Analyze Functions"

Problems for "Using the Second Derivative to Analyze Functions"

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.