1 Example 1 Explain why the Intermediate Value Theorem does or does not apply to each of the following functions. (a) f(x) = 1/x with domain [1,2]. Solution.

Slides:

Advertisements

Similar presentations

1 Example 3 (a) Find the range of f(x) = 1/x with domain [1,  ). Solution The function f is decreasing on the interval [1,  ) from its largest value.

Advertisements

LIMITS Continuity LIMITS In this section, we will: See that the mathematical definition of continuity corresponds closely with the meaning of the.

APPLICATIONS OF DIFFERENTIATION The Mean Value Theorem APPLICATIONS OF DIFFERENTIATION In this section, we will learn about: The significance of.

Ch 3.3: Linear Independence and the Wronskian

1 Example 1 (a) Let f be the rule which assigns each number to its square. Solution The rule f is given by the formula f(x) = x 2 for all numbers x. Hence.

Section 5.5 – The Real Zeros of a Rational Function

Rational Functions Find the Domain of a function

f /(x) = 6x2 – 6x – 12 = 6(x2 – x – 2) = 6(x-2)(x+1).

Continuity ( Section 1.8) Alex Karassev. Definition A function f is continuous at a number a if Thus, we can use direct substitution to compute the limit.

1.4 Continuity and One-Sided Limits This will test the “Limits” of your brain!

LIMITS 2. We noticed in Section 2.3 that the limit of a function as x approaches a can often be found simply by calculating the value of the function.

Section 1.4: Continuity and One-Sided Limits

Sec 5: Vertical Asymptotes & the Intermediate Value Theorem

Limit Laws Suppose that c is a constant and the limits lim f(x) and lim g(x) exist. Then x -> a Calculating Limits Using the Limit Laws.

Example Solution For a) ( f + g)(4)b) ( f – g)(x) c) ( f /g)(x)d) find the following. a) Since f (4) = 2(4) – (4) 2 = 8  16 =  8 and g(4) = 3(4)

Continuity and Discontinuity

Calculus, 8/E by Howard Anton, Irl Bivens, and Stephen Davis Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. 2.5 CONTINUITY Intuitively,

We noticed in Section 2.3 that the limit of a function as x approaches a can often be found simply by calculating the value of the function at a.  Functions.

Section 3.7 Proper Rational Functions Section 3.7 Proper Rational Functions.

Copyright © Cengage Learning. All rights reserved.

Asymptotes Objective: -Be able to find vertical and horizontal asymptotes.

Chapter 7 Polynomial and Rational Functions with Applications Section 7.2.

What is the symmetry? f(x)= x 3 –x.

Unit 1 Limits. Slide Limits Limit – Assume that a function f(x) is defined for all x near c (in some open interval containing c) but not necessarily.

Combined/Composite Function Continuity and the Intermediate Value Theorem Lesson

2.3 Continuity.

Review Limits When you see the words… This is what you think of doing…  f is continuous at x = a  Test each of the following 1.

1-4: Continuity and One-Sided Limits

1.4 Continuity  f is continuous at a if 1. is defined. 2. exists. 3.

1 Example 2 (a) Find the range of f(x) = x 2 +1 with domain [0,2]. Solution The function f is increasing on the interval [0,2] from its smallest value.

2.7: Continuity and the Intermediate Value Theorem Objectives: Define and explore properties of continuity Introduce Intermediate Value Theorem ©2002.

Sullivan PreCalculus Section 3.6 Real Zeros of a Polynomial Function Objectives Use the Remainder and Factor Theorems Use Descartes’ Rule of Signs Use.

Sullivan Algebra and Trigonometry: Section 5.2 Objectives Use the Remainder and Factor Theorems Use Descartes’ Rule of Signs Use the Rational Zeros Theorem.

4.2 – The Mean Value Theorem

Section 5.5 The Intermediate Value Theorem Rolle’s Theorem The Mean Value Theorem 3.6.

APPLICATIONS OF DIFFERENTIATION 4. We will see that many of the results of this chapter depend on one central fact—the Mean Value Theorem.

Intermediate Value Theorem Vince Varju. Definition The Intermediate Value Theorem states that if a function f is a continuous function on [a,b] then there.

On [a,b], ARC = On [1, 16], find ARC for. On [a,b], ARC = On [1, 16], find ARC for ARC = =

Section 3.6 Reciprocal Functions Section 3.6 Reciprocal Functions.

AP Calc AB IVT. Introduction Intermediate Value Theorem If a function is continuous between a and b, then it takes on every value between and. Because.

Exponential & Logarithmic functions. Exponential Functions y= a x ; 1 ≠ a > 0,that’s a is a positive fraction or a number greater than 1 Case(1): a >

Suppose you drive 200 miles, and it takes you 4 hours. Then your average speed is: If you look at your speedometer during this trip, it might read 65 mph.

Section 1-1: Relations and Functions *Relation: *Domain: *Range: *Function: Example 1: State the domain and range of each relation. Then state whether.

Discrete Mathematics Lecture # 17 Function. Relations and Functions  A function F from a set X to a set Y is a relation from X to Y that satisfies the.

Limits and Their Properties 1 Copyright © Cengage Learning. All rights reserved.

Copyright © Cengage Learning. All rights reserved.

4.2 The Mean Value Theorem In this section, we will learn about:

Rational Functions.

4.4 Rational Functions A Rational Function is a function whose rule is the quotient of two polynomials. i.e. f(x) = 1

Sullivan Algebra and Trigonometry: Section 5

4.1 Notes day 2 Remainder Theorem: If a polynomial f(x) is divided by x – c, then the remainder is f(c). Ex. f(x) = x3 + 3 divided by g(x)= x -1.

Properties of Functions

Functions JEOPARDY.

26 – Limits and Continuity II – Day 2 No Calculator

2.2 Limits Involving Infinity

Important Values for Continuous functions

Rational Functions, Transformations

Intermediate Value Theorem

2.5 Continuity In this section, we will:

Intermediate Value Theorem

Problem of the day.

4.6 The Mean Value Theorem.

8.2: Graph Simple Rational Functions

Continuity Alex Karassev.

1.4 Continuity and One-Sided Limits

The Intermediate Value Theorem

26 – Limits and Continuity II – Day 1 No Calculator

Ex1 Which mapping represents a function? Explain

1.4 Continuity and One-Sided Limits This will test the “Limits”

Presentation transcript:

1 Example 1 Explain why the Intermediate Value Theorem does or does not apply to each of the following functions. (a) f(x) = 1/x with domain [1,2]. Solution The function f is continuous on the interval [1,2]. Hence the Intermediate Value Theorem applies to f. It says that if A = f(a) = 1/a and B = f(b) = 1/b are selected with 1  a  b  2, then all numbers between A and B are in the range of f. That is, if B = 1/b  C  1/a = A then C = f(1/C). Moreover, since 1  a  1/C  b  2, the number 1/C is in the domain [1,2] of f.

2 (b) g(x) = 1/x with domain (0,1]. Solution The function g is continuous on the interval (0,1]. Hence the Intermediate Value Theorem applies to g. It says that if A = g(a) = 1/a and B = g(b) = 1/b are selected with 0 < a  b  1, then all numbers between A and B are in the range of g. That is, if B = 1/b  C  1/a = A then C = g(1/C). Moreover, since 0 < a  1/C  b  1, the number 1/C is in the domain (0,1] of g.

3 (c) h(x) = 1/x with domain (1,  ). Solution The function h is continuous on the interval (1,  ). Hence the Intermediate Value Theorem applies to h. It says that if A = h(a) = 1/a and B = h(b) = 1/b are selected with 1 < a  b, then all numbers between A and B are in the range of h. That is, if B = 1/b  C  1/a = A then C = h(1/C). Moreover, since 1 < a  1/C  b, the number 1/C is in the domain (1,  ) of h.

4 (d) k(x) = 1/x Solution The domain of k consists of all numbers for which the denominator x of 1/x is nonzero, i.e all numbers other than zero. Thus the domain of k is (- ,0)  (0,  ) which is not an interval. Hence the Intermediate Value Theorem does not apply to k. Note that the conclusion of the Intermediate Value Theorem is not valid for k: -1 = k(-1) < 0 < k(1) = 1 and –1, 1 are in the domain of k. However, 0 is not in the range of k - there is no number whose reciprocal equals zero. We can also see this from the graph of k below. Although the graph of k has height –1 at x=-1 and height +1 at x=+1, it never has height 0. The reason is apparent – the graph of k is not continuous at x=0 where it has a vertical asymptote.