# Objectives: 1.Be able to determine where a function is concave upward or concave downward with the use of calculus. 2.Be able to apply the second derivative.

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Objectives: 1.Be able to determine where a function is concave upward or concave downward with the use of calculus. 2.Be able to apply the second derivative test to find the relative extrema of a function. Critical Vocabulary: Concave Upward, Concave Downward Warm Up: Find the 2 nd derivative of each function

Let f be differentiable on a open interval I. The graph of f is concave upward on I if f’(x) is increasing on the interval and concave downward on I if f’(x) is decreasing on the interval.

(-∞,0) (0, ∞) This is called a point of inflection. Concavity changes at these points. (0, 2) (-∞, 0) (2, ∞)

Let f be a function whose second derivative exists on an open interval I. 1. If f’’(x) > 0 for all x in I, then f is Concave Upward in I 2. If f’’(x) < 0 for all x in I, then f is Concave Downward in I

Example 1: Find the open intervals on which the following function is concave upward or concave downward. Interval (-∞, -1)(-1, 1)(1, ∞) Test Valuex = -2x = 0x = 2 Sign of f’’(x) f’’(x) = (+)f’’(x) = (-)f’’(x) = (+) Conclusion Concave UpConcave DownConcave Up x = 1 and x = -1 Concave Up: Concave Down: (-∞, -1) (-1, 1) (1, ∞) These are called points of inflection. Concavity changes at these points.

Example 2: Find the open intervals on which the following function is concave upward or concave downward. Interval (-∞, -2)(-2, 2)(2, ∞) Test Valuex = -3x = 0x = 3 Sign of f’’(x) f’’(x) = (+)f’’(x) = (-)f’’(x) = (+) Conclusion Concave UpConcave DownConcave Up Discontinuity: x = +/-2 Concave Up: Concave Down: (-∞, - 2) (-2, 2) (2, ∞)

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Let f be a function such that f’(c) = 0 and the second derivative of f exists on a open interval containing c. 1.If f’’(c) > 0, then f(c) is a relative minimum of f. 2. If f’’(c) < 0, then f(c) is a relative maximum of f. 3. If f’’(c) = 0, then the test fails. In such case, you use the First Derivative Test.

Example 3: Find the relative extrema for f(x) = -3x 5 + 5x 3 using the second derivative test. f’(x) = -15x 4 + 15x 2 Critical Numbers: x = 0, 1, -1 f(0) = 0: (0, 0) f(1) = 2: (1, 2) f(-1) = -2: (-1, -2) f’’(x) = -60x 3 + 30x Point Sign f’’(x) Conclusion (-1, -2)(1, 2)(0, 0) f’’(-1) > 0 Relative Min f’’(1) < 0 Relative Max f’’(0) = 0 Test Fails Inflection Points: x = 0 Inflection

Example 4: Find the relative extrema for using the second derivative test. Critical Numbers: x = 0 f(0) = 1: (0, 1) Point Sign f’’(x) Conclusion (0, 1) f’’(0) > 0 Relative Min Inflection Points: None

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