Index FAQ

4.1 Intermediate Value Theorem for Continuous Functions

Review of Properties of Continuous FunctionsFunction that is Continuous at Irrational Points OnlyBolzano’s TheoremIntermediate Value Theorem for Continuous FunctionsApplications of the Intermediate Value Theorem

Index FAQMika Seppälä: Intermediate Value Theorem

Continuous Functions (1)

The function f : is in an interval if

it is continous at each point of the

c

i

ontinu

nte

ous

rval.

0

0

0

A function f : is at if the limit

li

continu

mf

ous

f x .x x

x x

x

Definition 1

Continuous function Discontinuous function

A function which is not continuous (at a point or in an interval) is said to be discontinuous.

Index FAQMika Seppälä: Intermediate Value Theorem

Continuous Functions (2)

0

0 0

Assume that f : is continuous at and that

f 0. Then 0 such that f 0.

x x

x x x x

0

0

Choose f in the definition of continuity.

We can do this since, by the assumption, f 0.

x

x

0 0 0

Hence

0 such that f f f .x x x x x

0

0 0

A function f : is at if

0 : 0 such tha

contin

t f

uo

f .

us x x

x x x x

Definition 2

Lemma

Proof

0 0 0 0 0 0f f f f f f f f fx x x x x x x x x

0f 2 f f 0.x x x

Index FAQMika Seppälä: Intermediate Value Theorem

Continuous Functions (3)

The following functions are continuous at all points where they take finite values.

1. Polynomials – they are continuous everywhere

2. Rational functions

3. Functions defined by algebraic expressions

4. Exponential functions and their inverses

5. Trigonometric functions and their inverses

Index FAQMika Seppälä: Intermediate Value Theorem

Continuous Functions (4)

Assume that f and g are continuous functions.

Theorem The following functions are continuous:

1. f + g

2. f g

3. f / g provided that g 0, i.e. the function is continuous at all points x for which g(x) 0.

Index FAQMika Seppälä: Intermediate Value Theorem

Continuous Functions (5)

0 0

0

Assume that f is continuous at g x , g continuous at

and that f g is defined. Then f g is continuous at .

x

x x

Lemma

Corollary

0 0 0

0 0

0

Assume that f is continuous at g x , g continuous at

and that f g is defined. Then

lim f g lim f g f lim g f g . x x x x x x

x

x x x x

Index FAQMika Seppälä: Intermediate Value Theorem

Continuous Functions (6)

Represent rational numbers as , , and

and do not have common factors other than 1.

mm n

nm n

Example Function which is continuous at irrational points and discontinuous at rational points.

0 if

1f if

1 if 0

x

mx x

n nx

Define

Index FAQMika Seppälä: Intermediate Value Theorem

Continuous Functions (7)

The function f defined in this way is clearly discontinuous at rational

1points since f 0 and there are, arbitrarily close to the

point , irrational points where f 0.

m m

n n n

mx x

n

0 if

1f if

1 if 0

x

mx x

n nx

Define

Continuity at irrational points follows from the fact that when approximating an irrational number by a rational number m/n, the denominator n grows arbitrarily large as the approximation gets better.

Index FAQMika Seppälä: Intermediate Value Theorem

Theorem for the existence of zeros (1)

Let , f 0 . M x a b x

Assume that the function f is continuous on an interval , ,

, and that f 0 and f 0. Then there is a point

, such that f 0.

a b

a b a b

a b

Since, by the assumptions, f 0, a ,

and hence .

a M

M

Bolzano’s Theorem

Proof

a

b

By the construction of the set , : .M x M x b

Hence is non-empty and bounded from the above.

This implies that sup exists and that , .

M

M a b

Index FAQMika Seppälä: Intermediate Value Theorem

Theorem for the existence of zeros (2)

Let sup sup , f 0 . Claim: f 0. M x a b x

To prove the claim assume that it is not true, i.e. assume that f 0.

If f 0, then 0 such that f 0.x x

Assume that the function f is continuous on an interval , , , and that

f 0 and f 0. Then there is a point , such that f 0.

a b a b

a b a b

Bolzano’sTheorem

Proof (cont’d)

This means that sup is not the upper bound

for the set , which is a contradiction.

small

Hence f 0

t

.

esM

M

If f 0, then we see, exactly in the same way, that is not an upper bound

for the set . This is also a contradiction. We conclude that f 0.M

Index FAQMika Seppälä: Intermediate Value Theorem

Intermediate Value Theorem for Continuous Functions

Assume that the function f is continuous on an interval , ,

, and that the number is between f and f .

Then there is a number between and such that f .

a b

a b c a b

a b c

Theorem

Proof If c > f(a), apply the previously shown Bolzano’s Theorem to the function f(x) - c.

Otherwise use the function c – f(x).

The Intermediate Value Theorem means that a function, continuous on an interval, takes any value between any two values that it takes on that interval. A continuous function cannot grow from being negative to positive without taking the value 0.

Index FAQMika Seppälä: Intermediate Value Theorem

Using the Intermediate Value Theorem (1)

Show that the equation cos 2 0 has a solution.

Find an approximation of the solution with error 0.001.

x x

Consider the function f cos 2 .

With this notation: cos 2 0 f 0.

x x x

x x x

The function f is continuous. Since f 0 1 and f 1 cos 1 2 0,

the Intermediate Value Theorem implies that there is 0,1 such that

f 0. Hence the equation has a solution in the interval 0,1 .

Problem

Solution

Index FAQMika Seppälä: Intermediate Value Theorem

Using the Intermediate Value Theorem (2) Show that the equation cos 2 0 has a solution.

Find an approximation of the solution with error 0.001.

x x

We now know that the solution for the equation

is in the interval 0,1 . To get a better approximation

of the solution , evaluate the function f at the midpoint

of the interval 0,1 , i.e. at the poi

1

nt .2

1 1Otherwise f 1 f 0, and there is a solution in the interval ,1 .

2 2

Problem

Solution

(cont’d)

1 1Assume that f 0. Then if f 0 f 0, we know, by the

2 2

1Intermediate Value Theorem, that there is a solution in the interval 0, .

2

Repeat the above to find an interval with length <0.002 containing the solution. The mid-point of this interval is the desired approximation.

Index FAQMika Seppälä: Intermediate Value Theorem

Using the Intermediate Value Theorem (3) Show that the equation cos 2 0 has a solution.

Find an approximation of the solution with error 0.001.

x x

Look at the function f cos 2 first in the interval 0,1 .

Make the interval smaller to approximate the solution. Repeat

this to get the desired accuracy.

x x x

Problem

GraphicalSolution

10

0.50

0.450180.450195

17th iteration, ξ0.450187501st iteration, ξ0.5 2nd iteration, ξ0.25