Download presentation

Presentation is loading. Please wait.

Published byJustyn Sawdon Modified over 3 years ago

1
1 CONCAVE UPWARDS g"(x) > 0

2
2 CONCAVE DOWNWARDS g"(x) < 0 negative slope y = g(x) positive slope zero slope

3
3 POINTS OF INFLECTION If a change in concavity occurs, then an inflection point exists. CU CD

4
4 EXAMPLE 1 Determine the maximum, minimum and inflection points of the function f(x) = x 3 – 3x 2 – 9x + 15 Step 1: Find the first and second derivatives f ′ (x) = 3x 2 – 6x – 9 f " (x) = 6x – 6 Step 2: Find critical values of x by setting the first derivative equal to 0. 3x 2 – 6x – 9 = 0 x 2 – 2x – 3 = 0 (x + 1) (x – 3) = 0 x = –1 or x = 3

5
5 Step 3: Use the second derivative to determine maximum or minimum f " (x) = 6x – 6 f " (–1) = 6(–1) – 6 = –12 f " (3) = 6(3) – 6 = 12 f(–1) = (–1) 3 – 3(–1) 2 – 9(–1) + 15 = 20 P(–1, 20) is a maximum point on the graph 20 is the maximum value f(3) = (3) 3 – 3(3) 2 – 9(3) + 15 = –12 P(3, –12) is a minimum point on the graph -12 is the minimum value

6
6 EXAMPLE 1 continued Step 4: Set the second derivative equal to zero and solve for the potential inflection point:. f " (x) = 6x – 6 6x – 6 = 0 x = 1 1 f " (0) = - 6 f " (2) = 6 CDCU Since the concavity changes at x = 1, we know that there is an inflection point at this value Step 5: Find the y value of the point. f(1) = (1) 3 – 3(1) 2 – 9(1) + 15 = 4 IP (1, 4)

7
7 (1, 4)

8
8 EXAMPLE 2 Find the concavity intervals and any points of inflection for the function Step 1: Find the first and second derivatives Step 2: Set the second derivative equal to zero and solve for the potential inflection points x = 0 or x = 2

9
9 Step 3: Test where the second derivative is positive or negative. 02 g"(-1) = 9g"(1) = -3g"(3) = 9 CUCDCU Since the concavity changes at both 0 and 2 there are inflection points at these x -values. g(0) = 5 and g(2) = -3 The infection points are at (0, 5) and (2, -3)

10
10 (0, 5) (2, -3)

11
11 EXAMPLE 3 Find the concavity intervals and any points of inflection for the function h(x) = x – cos x on the interval (0, 2 ) Step 1: Find the first and second derivatives Step 2: Set the second derivative equal to zero and solve for the potential inflection points h’(x) = 1 + sin x h“(x) = cos x cos x = 0 x = / 2 or x = 3 / 2

12
12 Step 3: Test where the second derivative is positive or negative. h“(x) = cos x /2/2 CUCD CU / 2 CAST Rule All Sin TanCos 0 22 + + Since the concavity changes at both / 2 and / 2 there are inflection points at these x -values. h( / 2 ) = / 2 and h( / 2 ) = / 2 The inflection points are at ( / 2, / 2 ) and ( / 2, / 2 )

13
13

Similar presentations

OK

BY DR. JULIA ARNOLD USING TAN’S 5TH EDITION APPLIED CALCULUS FOR THE MANAGERIAL, LIFE, AND SOCIAL SCIENCES TEXT Applications of the Second Derivative.

BY DR. JULIA ARNOLD USING TAN’S 5TH EDITION APPLIED CALCULUS FOR THE MANAGERIAL, LIFE, AND SOCIAL SCIENCES TEXT Applications of the Second Derivative.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on social networking sites in india Ppt on autism statistics Ppt on oxygen cycle and ozone layer Interactive ppt on male reproductive system Ppt on management by objectives mbo Ppt online templates Ppt on intellectual property rights Ppt on religion and science Ppt on centre of mass Ppt on mind reader computer