2.2 Geometrical Application of Calculus

Slides:

Advertisements

Similar presentations

Suppose that f(x) and g(x) are functions for which g(x) = f(x) – 5 for all values of x. a.How are the graphs of f(x) and g(x) related geometrically? b.

Advertisements

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.

Completing the square Expressing a quadratic function in the form:

The original function is… f(x) is… y is…

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.

f has a saddle point at (1,1) 2.f has a local minimum at.

1 f ’’(x) > 0 for all x in I f(x) concave Up Concavity Test Sec 4.3: Concavity and the Second Derivative Test the curve lies above the tangentsthe curve.

Increasing / Decreasing Test

CALCULUS I Chapter III Additionnal Applications of the Derivative Mr. Saâd BELKOUCH.

In this section, we will investigate some graphical relationships between a function and its second derivative.

CHAPTER Continuity Derivatives and the Shapes of Curves.

Section 5.3 – The Definite Integral

Curve Sketching 2.7 Geometrical Application of Calculus

Calculus Review. How do I know where f is increasing? O It is where f prime is positive. O Find the derivative and set equal to zero. Use test points.

Differentiation. f(x) = x 3 + 2x 2 – 3x + 5 f’(x) = 3x 2 + 4x - 3 f’(1) = 3 x x 1 – 3 = – 3 = 4 If f(x) = x 3 + 2x 2 –

Increasing/ Decreasing

Aim: Curve Sketching Do Now: Worksheet Aim: Curve Sketching.

Copyright © Cengage Learning. All rights reserved. Differentiation 2.

Analyzing Graphs of Polynomial Functions. With two other people: Each person pick a letter, f, g, or h Each person will graph their function After graphing.

Objectives Use the first-order derivative to find the stationary points of a function. Use the second-order derivative to classify the stationary points.

Unit 1 Review Standards 1-8. Standard 1: Describe subsets of real numbers.

Section 3.5 Summary of Curve Sketching. THINGS TO CONSIDER BEFORE SKETCHING A CURVE Domain Intercepts Symmetry - even, odd, periodic. Asymptotes - vertical,

Definition of Curve Sketching  Curve Sketching is the process of using the first and second derivative and information gathered from the original equation.

3.9: Derivatives of Exponential and Logarithmic Functions.

Warmup  If y varies directly as the square of x and inversely as z and y =36 when x =12 and z =8, find x when y=4, and z=32.

5.3 Definite Integrals and Riemann Sums. I. Rules for Definite Integrals.

Calculus Continued Tangents and Normals Example Find the equations of the tangent and normal to the graph of at the point where.

Stationary/Turning Points How do we find them?. What are they?  Turning points are points where a graph is changing direction  Stationary points are.

§3.4 Concavity Concave Up Concave Down Inflection Points Concavity Changes Concave Up Concave Down.

AP Calculus Chapter 5. Definition Let f be defined on an interval, and let x 1 and x 2 denote numbers in that interval f is increasing on the interval.

4.3 – Derivatives and the shapes of curves

AP Calculus AB Extrema, Inflection Points, and Propagated Error Calculator.

First derivative: is positive Curve is rising. is negative Curve is falling. is zero Possible local maximum or minimum. Second derivative: is positive.

OPTIMIZATION IN BUSINESS/ECONOMICS

Relative Extrema and More Analysis of Functions

LCHL Strand 5 Functions/Calculus Stationary Points

Applications of Differential Calculus DP Objectives: 7.4, 7.5, 7.6

2-5 Absolute Value Functions and Graphs

3.6 Critical Points and Extrema

Title: Higher order Derivatives & Stationary point H/W Marking

Lesson 13: Analyzing Other Types of Functions

3-6 Critical Points and Extrema

Applications of Differential Calculus

Title: Maximum, Minimum & Point of inflection

Lesson 13: Analyzing Other Types of Functions

Concavity and Second Derivative Test

A function f is increasing on an open interval I if, for any choice of x1 and x2 in I, with x1 < x2, we have f(x1) < f(x2). A function f is decreasing.

Lesson 11: Exponential Functions

Second Derivative Test

Concavity and the Second Derivative Test

1 2 Sec 4.3: Concavity and the Second Derivative Test

The 2nd Derivative.

Application of Derivative in Analyzing the Properties of Functions

-20 is an absolute minimum 6 is an absolute minimum

3.6 – Critical Points & Extrema

MATH 1311 Section 1.3.

Section 6 – Critical Points and Extrema

Nuffield Free-Standing Mathematics Activity

WHAT YOU SHOULD HAVE ON YOUR DESK…

1 2 Sec4.3: HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH Concavity Test

MATH 1311 Section 1.3.

X y y = x2 - 3x Solutions of y = x2 - 3x y x –1 5 –2 –3 6 y = x2-3x.

What is the function of the graph? {applet}

1 2 Sec4.3: HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH

The First Derivative Test

Math 1304 Calculus I 4.03 – Curve Shape.

- Derivatives and the shapes of graphs - Curve Sketching

Presentation transcript:

2.2 Geometrical Application of Calculus Types of Stationary Points 1. Minimum - + - + - + - + - + - - + + x LHS Minimum RHS f’(x) < 0 = 0 > 0

2.2 Geometrical Application of Calculus Types of Stationary Points 2. Maximum - + - + - + - - + - + x LHS Maximum RHS f’(x) > 0 = 0 < 0

2.2 Geometrical Application of Calculus Types of Stationary Points - + 3.Point of Horizontal Inflection - + - + - - + - + - + - - + - + - x LHS Inflection RHS f’(x) > 0 < 0

2.2 Geometrical Application of Calculus Types of Stationary Points Find any stationary points on the curve f(x) = x2 - 2x & determine what type it is. f’(x) = 2x - 2 f(1) = (1)2 – 2(1) = -1 2x - 2 = 0 (Stationary) Stationary @ (1, -1) 2x = 2 f’(0) = 2(0) - 2 = -2 x = 1 f’(2) = 2(2) - 2 = +2 LHS x = 0 Stationary x = 1 RHS x = 2 - + (1, -1) is a Minimum

(0, 2) is a horizontal point of inflection 2.2 Geometrical Application of Calculus Types of Stationary Points 2. Find the turning point on the curve y = 2x3 + 2 and determine what type it is. f(x) = 2x3 + 2 f(0) = 2(0)3 + 2 = 2 f’(x) = 6x2 Stationary @ (0, 2) f’(-1) = 6(-1)2 = +6 6x2 = 0 (Stationary) x = 0 f’(1) = 6(1)2 = +6 LHS x = -1 Stationary x = 0 RHS x = 1 + (0, 2) is a horizontal point of inflection