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

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

Solution for Problem 2 >>

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.

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

Solution for Problem 3 >>

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.

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

Solution for Problem 4 >>

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.

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