Download presentation

Presentation is loading. Please wait.

Published byApril Leech Modified over 2 years ago

1
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

2
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.

3
(-∞,0) (0, ∞) This is called a point of inflection. Concavity changes at these points. (0, 2) (-∞, 0) (2, ∞)

4
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

5
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.

6
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, ∞)

7
Page 342 #1-9 odds (MUST USE CALCULUS!!!!)

8
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.

9
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

10
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

11
Page 342 #11-33 odd, 45, 47 (MUST USE CALCULUS!!!!)

Similar presentations

OK

Miss Battaglia AP Calculus AB/BC. Let f be differentiable on an open interval I. The graph of f is concave upward on I if f’ is increasing on the interval.

Miss Battaglia AP Calculus AB/BC. Let f be differentiable on an open interval I. The graph of f is concave upward on I if f’ is increasing on the interval.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on eisenmenger syndrome and pulmonary Ppt on classical economics graph Ppt on chromosomes and genes diagram Ppt online examination project in java Ppt on self development tips Ppt on model view controller diagram Ppt on total parenteral nutrition Ppt on event driven programming pdf Ppt on centring club Introduction to liquid crystal display ppt on tv