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

Slides:



Advertisements
Similar presentations
Application of Derivative in Analyzing the Properties of Functions
Advertisements

The Parabola 3.6 Chapter 3 Conics 3.6.1
First Derivative Test, Concavity, Points of Inflection Section 4.3a.
§ 2.3 The First and Second Derivative Tests and Curve Sketching.
§ 4.4 The Natural Logarithm Function.
Chapter 2 Applications of the Derivative
Addition 1’s to 20.
Unit 6 Lesson #1 Intercepts and Symmetry
Maximum ??? Minimum??? How can we tell?
U1A L6 Linear, Quadratic & Polynomial Inequalities
DO NOW: Find where the function f(x) = 3x4 – 4x3 – 12x2 + 5
Relationship between First Derivative, Second Derivative and the Shape of a Graph 3.3.
Section 3.4 – Concavity and the Second Derivative Test
Concavity and the Second Derivative Test
4.3 Derivatives and the shapes of graphs 4.4 Curve Sketching
Concavity & the second derivative test (3.4) December 4th, 2012.
Extremum & Inflection Finding and Confirming the Points of Extremum & Inflection.
Section 3.3 How Derivatives Affect the Shape of a Graph.
Concavity and Inflection Points The second derivative will show where a function is concave up or concave down. It is also used to locate inflection points.
Maximum and Minimum Value Problems By: Rakesh Biswas
Relating Graphs of f and f’
2.3 Curve Sketching (Introduction). We have four main steps for sketching curves: 1.Starting with f(x), compute f’(x) and f’’(x). 2.Locate all relative.
Sec 3.4: Concavity and the Second Derivative Test
Inflection Points. Objectives Students will be able to Determine the intervals where a function is concave up and the intervals where a function is concave.
Relative Extrema.
This is the graph of y = sin xo
Who wants to be a Millionaire? Hosted by Kenny, Katie, Josh and Mike.
First and Second Derivative Test for Relative Extrema
Extremum & Inflection. Finding and Confirming the Points of Extremum & Inflection.
Section 4.1 Using First and Second Derivatives. Let’s see what we remember about derivatives of a function and its graph –If f’ > 0 on an interval than.
Concavity f is concave up if f’ is increasing on an open interval. f is concave down if f’ is decreasing on an open interval.
Lesson 4-3 First and Second Derivative Test for Relative Extrema.
Section 5.2 – Applications of the Second Derivative.
Concavity and the Second- Derivative Test. 1. Determine the open intervals on which the graph of the function is concave upward or concave downward (similar.
The Shape of the Graph 3.3. Definition: Increasing Functions, Decreasing Functions Let f be a function defined on an interval I. Then, 1.f increases on.
Review Derivatives When you see the words… This is what you know…  f has a local (relative) minimum at x = a  f(a) is less than or equal to every other.
Concavity of a graph A function is concave upward on an interval (a, b) if the graph of the function lies above its tangent lines at each point of (a,
4.3 How Derivatives Affect the Shape of a Graph. Facts If f ’( x ) > 0 on an interval ( a,b ), then f (x) is increasing on ( a,b ). If f ’( x ) < 0 on.
Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:
2.1.  Remember that the first derivative test will tell where a function is increasing or decreasing.
76.8 – Average Rate of Change = = -9 = – Average Rate of Change = = -9 =
Applications of Derivatives
CHAPTER Continuity Derivatives and the Shapes of Curves.
In the past, one of the important uses of derivatives was as an aid in curve sketching. We usually use a calculator of computer to draw complicated graphs,
Copyright © 2016, 2012 Pearson Education, Inc
Definition of Curve Sketching  Curve Sketching is the process of using the first and second derivative and information gathered from the original equation.
Ch. 5 – Applications of Derivatives
Copyright © Cengage Learning. All rights reserved. 3 Applications of Differentiation.
Extremum & Inflection. Finding and Confirming the Points of Extremum & Inflection.
Sketching Functions We are now going to use the concepts in the previous sections to sketch a function, find all max and min ( relative and absolute ),
If f(x) is a continuous function on a closed interval x ∈ [a,b], then f(x) will have both an Absolute Maximum value and an Absolute Minimum value in the.
First derivative: is positive Curve is rising. is negative Curve is falling. is zero Possible local maximum or minimum. Second derivative: is positive.
12.2 Second Derivative and Graphs
Ch. 5 – Applications of Derivatives
4.3 Using Derivatives for Curve Sketching.
Extreme Values of Functions
Review Problems Sections 3-1 to 3-4
4.3 Derivatives and the shapes of graphs 4.5 Curve Sketching
Second Derivative Test
Second Derivative Test
Sec 3.4: Concavity and the Second Derivative Test
3.4: Concavity and the Second Derivative Test
4.3 Connecting f’ and f’’ with the graph of f
Derivatives and Graphing
1 2 Sec4.3: HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH
Copyright © Cengage Learning. All rights reserved.
Concavity & the second derivative test (3.4)
Math 1304 Calculus I 4.03 – Curve Shape.
- Derivatives and the shapes of graphs - Curve Sketching
Presentation transcript:

1 CONCAVE UPWARDS g"(x) > 0

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

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

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 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 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 (1, 4)

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 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 (0, 5) (2, -3)

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 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 22 + + 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