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.

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

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.

Section 5.5 – The Real Zeros of a Rational Function

2.5 Descartes’ Rule of Signs To apply theorems about the zeros of polynomial functions To approximate zeros of polynomial functions.

1.Definition of a function 2.Finding function values 3.Using the vertical line test.

Copyright © 2011 Pearson, Inc. 2.3 Polynomial Functions of Higher Degree with Modeling.

Rolle’s theorem and Mean Value Theorem ( Section 3.2) Alex Karassev.

AP Calculus 1004 Continuity (2.3). C CONVERSATION: Voice level 0. No talking! H HELP: Raise your hand and wait to be called on. A ACTIVITY: Whole class.

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.

Intermediate Value Theorem If f is continuous on [ a,b ] and k is any number between f(a) and f(b), inclusive, then there is at least one number c in the.

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

Section 1.4: Continuity and One-Sided Limits

Sec 5: Vertical Asymptotes & the Intermediate Value Theorem

MAT 1234 Calculus I Section 1.8 Continuity

Graphs of Polynomial Functions

Warm up 1. Do in notebook. Be seated before the bell rings DESK homework Warm-up (in your notes) Agenda : warm-up Go over homework homework quiz Notes.

CONTINUITY Mrs. Erickson Continuity lim f(x) = f(c) at every point c in its domain. To be continuous, lim f(x) = lim f(x) = lim f(c) x  c+x  c+ x 

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.

Section 5.3(d) Synthetic Substitution. Long division Synthetic Division can be used to find the value of a function. This process is called Synthetic.

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

Polynomial Functions of Higher Degree. Quick Review.

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

Section 1.4 – Continuity and One-Sided Limits

Chapter 4 Polynomial (Poly) & Rational Functions Copyright ©2013, 2009, 2006, 2005 Pearson Education, Inc.

CONTINUITY. A function f(x) is continuous at a number a if: 3 REQUIREMENTS f(a) exists – a is in the domain of f exists.

Limits and Continuity Unit 1 Day 4.

Continuity and One- Sided Limits (1.4) September 26th, 2012.

Section 4.2 Graphing Polynomial Functions Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

4.2 Critical Points Mon Oct 19 Do Now Find the derivative of each 1) 2)

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.

1.4 Continuity and One-Sided Limits Main Ideas Determine continuity at a point and continuity on an open interval. Determine one-sided limits and continuity.

Algebra and Calculus 8-1 Copyright © Genetic Computer School 2007 Lesson 8 Fundamentals of Calculus.

Divide by x - 1 Synthetic Division: a much faster way!

3.2 Rolle’s Theorem and the

Rolle’s theorem and Mean Value Theorem (Section 4.2)

Polynomial Functions of Higher Degree with Modeling

3.7 The Real Zeros of a Polynomial Function

Locating Real Zeros of Polynomials

Table of Contents 8. Section 2.7 Intermediate Value Theorem.

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.

2.2 Polynomial Function of Higher Degrees

Tangent Lines and Theorems Notes

3.7 The Real Zeros of a Polynomial Function

3.2 Rolle’s Theorem and the

Graphs of Polynomial Functions

AP Calculus September 6, 2016 Mrs. Agnew

Important Values for Continuous functions

Intermediate Value Theorem

1. Be able to apply The Mean Value Theorem to various functions.

Intermediate Value Theorem

Table of Contents 8. Section 2.7 Intermediate Value Theorem.

1.4 Continuity and One-Sided Limits (Part 2)

Continuity Alex Karassev.

Warm-up: Determine the left and right-hand behavior of the graph of the polynomial function, then find the x-intercepts (zeros). y = x3 + 2x2 – 8x HW:

The Intermediate Value Theorem

Polynomial Functions of Higher Degree

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

Warm-up How are the concepts of a limit and continuity related?

A function f is continuous at the point x = a if the following are true: f(a) a.

5. Continuity on an Interval

Continuity and One-Sided limits

Lesson 63 - Intermediate Value Theorem

Maximum and Minimum Values

Presentation transcript:

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 exists a number c in (a,b) so that f(c) = n n - any number between f(a) and f(b) where f(a) ≠ f(b) IVT is used to find solutions or zeros in polynomials along a given Domain by testing points and determining if the the sign changes between those two points. f = x^2-2 test if there is a zero between (0,4) f(0) = -2 f(4) = 14 since there is a sign change, by the definition of IVT, there is a zero between 0 and 4.

Examples 1.f(x) = x^3 - 2x^2 + 15x - 10 determine if there is a zero between (0,2) f(0) = (0)^3 - 2(0)^2 + 15(0) - 10 = -10 f(2) = (2)^3-2(2)^2+15(2)-10 = = 20 Because the sign changes along the Domain of (0,2), by the definition of IVT, there is a zero between 0 and f(x) = x^4-x+2 determine if there is a zero between (3,5) f(3) = (3)^4-3+2 = 81-1 = 80 f(5) = (5)^4-5+2 = = 622 Because there is no sign change along the Domain of (3,5), there is no zero between 3 and 5.

Practice Problems 1. f(x) = x^5 - 5x^4 + 2x^3 - 6x^ determine if there is a zero between 0 and 4 2. f(x) = x^3 - 7x^2 + 10x - 4 determine if there is a zero between 1 and 5 3. f(x) = -5x^2 + 8x + 2 determine if there is a zero between 0 and 1

Solutions 1. f(0) = 15 f(4) = -209 by the definition of IVT, there is a zero between 0 and f(1) = 0 f(5) = -4 by the definition of IVT, there is a zero between 1 and f(0) = 2 f(2) = 5 by the definition of IVT, there is not a solution between 0 and 1.