MAT01A1: Continuity (and the IntermediateValue Theorem)

Dr Craig

17–20 March 2020

This lecture:

I Continuity

I Intermediate Value Theorem

The definition that appears on the next slide

is extremely important.

Are you ready?

OK. Here it is:

The definition that appears on the next slide

is extremely important.

Are you ready?

OK. Here it is:

The definition that appears on the next slide

is extremely important.

Are you ready?

OK. Here it is:

Definition: A function f is continuous at a

number a ∈ R if limx→a

f (x) = f (a).

Very important: the above definition implies

there are three conditions that must be

satisfied for a function to be continuous at a:

(1) f must be defined at a

(2) limx→a

f (x) must exist

(3) limx→a

f (x) = f (a)

Definition: A function f is continuous at a

number a ∈ R if limx→a

f (x) = f (a).

Very important: the above definition implies

there are three conditions that must be

satisfied for a function to be continuous at a:

(1) f must be defined at a

(2) limx→a

f (x) must exist

(3) limx→a

f (x) = f (a)

Definition: A function f is continuous at a

number a ∈ R if limx→a

f (x) = f (a).

Very important: the above definition implies

there are three conditions that must be

satisfied for a function to be continuous at a:

(1) f must be defined at a

(2) limx→a

f (x) must exist

(3) limx→a

f (x) = f (a)

Definition: A function f is continuous at a

number a ∈ R if limx→a

f (x) = f (a).

Very important: the above definition implies

there are three conditions that must be

satisfied for a function to be continuous at a:

(1) f must be defined at a

(2) limx→a

f (x) must exist

(3) limx→a

f (x) = f (a)

Definition: A function f is continuous at a

number a ∈ R if limx→a

f (x) = f (a).

Very important: the above definition implies

there are three conditions that must be

satisfied for a function to be continuous at a:

(1) f must be defined at a

(2) limx→a

f (x) must exist

(3) limx→a

f (x) = f (a)

If a function f (x) is not continuous when

x = a, we say that “f is discontinuous at a”.

A point at which a function is discontinuous

is called a discontinuity.

Examples of discontinuities

Recall that in order for f to be continuous at x = athere are three conditions that must be satisfied:

(1) f must be defined at a

(2) limx→a

f(x) must exist

(3) limx→a

f(x) = f(a)

(1) f is discontinuous at x = 1 because it is notdefined there.

(2) f is discontinuous at x = 3 becauselimx→3− f(x) 6= limx→3+ f(x).

(3) f is discontinuous at x = 5 becauselimx→5 f(x) 6= f(5)

(1) f is discontinuous at x = 1 because it is notdefined there.

(2) f is discontinuous at x = 3 becauselimx→3− f(x) 6= limx→3+ f(x).

(3) f is discontinuous at x = 5 becauselimx→5 f(x) 6= f(5)

(1) f is discontinuous at x = 1 because it is notdefined there.

(2) f is discontinuous at x = 3 becauselimx→3− f(x) 6= limx→3+ f(x).

(3) f is discontinuous at x = 5 becauselimx→5 f(x) 6= f(5)

More examples of discontinuities

The function above is discontinuous at

x = 0. This kind of discontinuity is know as

an infinite discontinuity.

Examples:

Where are the following discontinuous?

(a) f (x) =x2 − x− 2

x− 2

(b) f (x) =

{1x2

if x 6= 0

1 if x = 0

(c) f (x) =

{x2−x−2x−2 if x 6= 2

1 if x = 2

(a) f (x) =x2 − x− 2

x− 2Clearly f is undefined at x = 2 so it can’t

be continuous there.

Later we will have a

theorem that will tell us that f is

continuous at all x 6= 2.

(b) limx→0

f (x) =∞ and therefore limx→0

f (x)

does not exist. Hence f is not continuous

at x = 0.

(a) f (x) =x2 − x− 2

x− 2Clearly f is undefined at x = 2 so it can’t

be continuous there. Later we will have a

theorem that will tell us that f is

continuous at all x 6= 2.

(b) limx→0

f (x) =∞ and therefore limx→0

f (x)

does not exist. Hence f is not continuous

at x = 0.

(c) f(x) =

{x2−x−2x−2 if x 6= 2

1 if x = 2

Clearly f(2) is defined.

Now let us check iflimx→2 f(x) exists. We have:

limx→2

f(x) = limx→2

x2 − x− 2

x− 2

= limx→2

(x− 2)(x+ 1)

x− 2= lim

x→2(x+ 1) = 3.

So, limx→2 f(x) exists, but limx→2 f(x) 6= f(2),so f is not continuous at x = 2.

(c) f(x) =

{x2−x−2x−2 if x 6= 2

1 if x = 2

Clearly f(2) is defined. Now let us check iflimx→2 f(x) exists.

We have:

limx→2

f(x) = limx→2

x2 − x− 2

x− 2

= limx→2

(x− 2)(x+ 1)

x− 2= lim

x→2(x+ 1) = 3.

So, limx→2 f(x) exists, but limx→2 f(x) 6= f(2),so f is not continuous at x = 2.

(c) f(x) =

{x2−x−2x−2 if x 6= 2

1 if x = 2

Clearly f(2) is defined. Now let us check iflimx→2 f(x) exists. We have:

limx→2

f(x) = limx→2

x2 − x− 2

x− 2

= limx→2

(x− 2)(x+ 1)

x− 2= lim

x→2(x+ 1) = 3.

So, limx→2 f(x) exists, but limx→2 f(x) 6= f(2),so f is not continuous at x = 2.

(c) f(x) =

{x2−x−2x−2 if x 6= 2

1 if x = 2

Clearly f(2) is defined. Now let us check iflimx→2 f(x) exists. We have:

limx→2

f(x) = limx→2

x2 − x− 2

x− 2

= limx→2

(x− 2)(x+ 1)

x− 2= lim

x→2(x+ 1) = 3.

So, limx→2 f(x) exists, but limx→2 f(x) 6= f(2),so f is not continuous at x = 2.

Types of discontinuities

In example (a) and example (c) we have a

removable discontinuity. We call them

“removable” because if we redefined that

point we could “remove” the discontinuity.

In both of those examples we could remove

the discontinuity if we defined f (2) = 3.

The discontinuity in (b) is called an infinitediscontinuity.

Types of discontinuities

In example (a) and example (c) we have a

removable discontinuity. We call them

“removable” because if we redefined that

point we could “remove” the discontinuity.

In both of those examples we could remove

the discontinuity if we defined f (2) = 3.

The discontinuity in (b) is called an infinitediscontinuity.

Types of discontinuities

In example (a) and example (c) we have a

removable discontinuity. We call them

“removable” because if we redefined that

point we could “remove” the discontinuity.

In both of those examples we could remove

the discontinuity if we defined f (2) = 3.

The discontinuity in (b) is called an infinitediscontinuity.

Another example (d)

Consider the greatest integer function JxK.

JxK = the largest integer that is less than or

equal to x

Examples:

J0.5K = 0 J−1.2K = −2 J3K = 3

Where is this function discontinuous?

Greatest integer function

f (x) = JxK

Example (d) has jump discontinuities.

One-sided continuity:

Definition: A function f is continuous

from the right at a if

limx→a+

f (x) = f (a)

and continuous from the left at a if

limx→a−

f (x) = f (a)

Example of one-sided continuity:

The Heaviside function:

H(t) =

{0 if t < 0

1 if t > 0

The function is continuous from the right at

0, but not continuous from the left at 0.

This is because

limt→0+

H(t) = 1 = H(0)

but

limt→0−

H(t) = 0 6= H(0)

Example of one-sided continuity:

The Heaviside function:

H(t) =

{0 if t < 0

1 if t > 0

The function is continuous from the right at

0, but not continuous from the left at 0.

This is because

limt→0+

H(t) = 1 = H(0)

but

limt→0−

H(t) = 0 6= H(0)

Definition:

A function f is continuous on an intervalif it is continuous at every number in the

interval.

Note: if f is defined only on one side of an

endpoint of the interval, we understand

“continuous” to mean “continuous from the

right” or “continuous from the left”.

An example of this last point is f (x) =√x.

This function is continuous on [0,∞).

Definition:

A function f is continuous on an intervalif it is continuous at every number in the

interval.

Note: if f is defined only on one side of an

endpoint of the interval, we understand

“continuous” to mean “continuous from the

right” or “continuous from the left”.

An example of this last point is f (x) =√x.

This function is continuous on [0,∞).

Definition:

A function f is continuous on an intervalif it is continuous at every number in the

interval.

Note: if f is defined only on one side of an

endpoint of the interval, we understand

“continuous” to mean “continuous from the

right” or “continuous from the left”.

An example of this last point is f (x) =√x.

This function is continuous on [0,∞).

Example:

Show that f (x) = 1−√1− x2 is

continuous on the interval [−1, 1].

To do this we must prove three equalities:

I limx→a

f (x) = f (a) for any a ∈ (−1, 1)I lim

x→−1+f (x) = f (−1)

I limx→1−

f (x) = f (1).

Example:

Show that f (x) = 1−√1− x2 is

continuous on the interval [−1, 1].

To do this we must prove three equalities:

I limx→a

f (x) = f (a) for any a ∈ (−1, 1)I lim

x→−1+f (x) = f (−1)

I limx→1−

f (x) = f (1).

The graph of f (x) = 1−√1− x2 is the

lower half of the circle x2 + (y − 1)2 = 1, i.e.

the lower half of the circle of radius 1

centred at the point (0; 1).

Building continuous functions:

Theorem: if f and g are continuous at a

and c is a constant, then the following

functions are also continuous at a:

1. f + g

2. f − g3. cf

4. fg

5.f

gif g(a) 6= 0

Proving (1):

If f and g are continuous at a then the

function

f + g

is also continuous at a.

Q: How do we prove this?

A: Using Limit Laws and the definition of

the function f + g. In particular we use the

fact that the limit of a sum is the sum of the

limits (provided that both limits exist).

Proving (1):

If f and g are continuous at a then the

function

f + g

is also continuous at a.

Q: How do we prove this?

A: Using Limit Laws and the definition of

the function f + g. In particular we use the

fact that the limit of a sum is the sum of the

limits (provided that both limits exist).

Proving (1):

If f and g are continuous at a then the

function

f + g

is also continuous at a.

Q: How do we prove this?

A: Using Limit Laws and the definition of

the function f + g. In particular we use the

fact that the limit of a sum is the sum of the

limits (provided that both limits exist).

The next two theorems are very important.

Part of the reason for their importance is

that they make our lives a lot easier. They

tell us that a large number of functions are

continuous everywhere on their domains.

Theorem:

(a) Any polynomial is continuous

everywhere; that is, it is continuous on

R = (−∞,∞).

(b) Any rational function is continuous

wherever it is defined; that is, it is

continuous on its domain.

Theorem: the following types of

functions are continuous at every number

in their domains:

I polynomials

I rational functions

I root functions

I trig functions

I inverse trig functions

I exponential functions

I logarithmic functions

Example:

Where and why is the following function

continuous:

f (x) =`n(x) + arctan(x)

x2 − 1

Attempt this on your own and then look at

the solution on the next slide. Hint: think of

the domains of each component function and

the domain of f (x).

Example:

Where and why is the following function

continuous:

f (x) =`n(x) + arctan(x)

x2 − 1Attempt this on your own and then look at

the solution on the next slide. Hint: think of

the domains of each component function and

the domain of f (x).

Where and why is the following function continuous:

f(x) =`n(x) + arctan(x)

x2 − 1Solution:

We have `n(x) is continuous everywhereon its domain because it is a log function. That is,it is continuous on (0,∞).

Now dom(arctanx) = (−∞,∞) and it is an inversetrig function, hence it is continuous everywhere onits domain. Hence `n(x) + arctanx is continuouson (0,∞). (Recall: dom(f + g) = dom f ∩ dom g.)

Lastly, the denominator is continuous everywhere asit is a polynomial. However, f(x) is not defined forx = ±1 so it will not be continuous at those points.

Therefore f(x) will be continuous on (0, 1)∪ (1,∞).

Where and why is the following function continuous:

f(x) =`n(x) + arctan(x)

x2 − 1Solution: We have `n(x) is continuous everywhereon its domain because it is a log function. That is,it is continuous on (0,∞).

Now dom(arctanx) = (−∞,∞) and it is an inversetrig function, hence it is continuous everywhere onits domain. Hence `n(x) + arctanx is continuouson (0,∞). (Recall: dom(f + g) = dom f ∩ dom g.)

Lastly, the denominator is continuous everywhere asit is a polynomial. However, f(x) is not defined forx = ±1 so it will not be continuous at those points.

Therefore f(x) will be continuous on (0, 1)∪ (1,∞).

Where and why is the following function continuous:

f(x) =`n(x) + arctan(x)

x2 − 1Solution: We have `n(x) is continuous everywhereon its domain because it is a log function. That is,it is continuous on (0,∞).

Now dom(arctanx) = (−∞,∞) and it is an inversetrig function, hence it is continuous everywhere onits domain.

Hence `n(x) + arctanx is continuouson (0,∞). (Recall: dom(f + g) = dom f ∩ dom g.)

Lastly, the denominator is continuous everywhere asit is a polynomial. However, f(x) is not defined forx = ±1 so it will not be continuous at those points.

Therefore f(x) will be continuous on (0, 1)∪ (1,∞).

Where and why is the following function continuous:

f(x) =`n(x) + arctan(x)

x2 − 1Solution: We have `n(x) is continuous everywhereon its domain because it is a log function. That is,it is continuous on (0,∞).

Now dom(arctanx) = (−∞,∞) and it is an inversetrig function, hence it is continuous everywhere onits domain. Hence `n(x) + arctanx is continuouson (0,∞).

(Recall: dom(f + g) = dom f ∩ dom g.)

Lastly, the denominator is continuous everywhere asit is a polynomial. However, f(x) is not defined forx = ±1 so it will not be continuous at those points.

Therefore f(x) will be continuous on (0, 1)∪ (1,∞).

Where and why is the following function continuous:

f(x) =`n(x) + arctan(x)

x2 − 1Solution: We have `n(x) is continuous everywhereon its domain because it is a log function. That is,it is continuous on (0,∞).

Now dom(arctanx) = (−∞,∞) and it is an inversetrig function, hence it is continuous everywhere onits domain. Hence `n(x) + arctanx is continuouson (0,∞). (Recall: dom(f + g) = dom f ∩ dom g.)

Lastly, the denominator is continuous everywhere asit is a polynomial. However, f(x) is not defined forx = ±1 so it will not be continuous at those points.

Therefore f(x) will be continuous on (0, 1)∪ (1,∞).

Where and why is the following function continuous:

f(x) =`n(x) + arctan(x)

x2 − 1Solution: We have `n(x) is continuous everywhereon its domain because it is a log function. That is,it is continuous on (0,∞).

Now dom(arctanx) = (−∞,∞) and it is an inversetrig function, hence it is continuous everywhere onits domain. Hence `n(x) + arctanx is continuouson (0,∞). (Recall: dom(f + g) = dom f ∩ dom g.)

Lastly, the denominator is continuous everywhere asit is a polynomial. However, f(x) is not defined forx = ±1 so it will not be continuous at those points.

Therefore f(x) will be continuous on (0, 1)∪ (1,∞).

Where and why is the following function continuous:

f(x) =`n(x) + arctan(x)

x2 − 1Solution: We have `n(x) is continuous everywhereon its domain because it is a log function. That is,it is continuous on (0,∞).

Now dom(arctanx) = (−∞,∞) and it is an inversetrig function, hence it is continuous everywhere onits domain. Hence `n(x) + arctanx is continuouson (0,∞). (Recall: dom(f + g) = dom f ∩ dom g.)

Lastly, the denominator is continuous everywhere asit is a polynomial. However, f(x) is not defined forx = ±1 so it will not be continuous at those points.

Therefore f(x) will be continuous on (0, 1)∪ (1,∞).

Example:

f (x) =

2x + 8a if x < 0

ax2 − 4b if 0 6 x 6 2

−3x + 18 if x > 2

What values of a and b make f continuous

everywhere?

Example:

f(x) =

2x+ 8a if x < 0

ax2 − 4b if 0 6 x 6 2

−3x+ 18 if x > 2

What values of a and b make f continuouseverywhere?

Hint: because each piece of the function is apolynomial, you know it is continuous on(−∞, 0) ∪ (0, 2) ∪ (2,∞). So, you just need valuesof a and b that will give us

limx→0

f(x) = 0 and limx→2

f(x) = 2.

Example:

f(x) =

2x+ 8a if x < 0

ax2 − 4b if 0 6 x 6 2

−3x+ 18 if x > 2

What values of a and b make f continuouseverywhere?

Hint: because each piece of the function is apolynomial, you know it is continuous on(−∞, 0) ∪ (0, 2) ∪ (2,∞).

So, you just need valuesof a and b that will give us

limx→0

f(x) = 0 and limx→2

f(x) = 2.

Example:

f(x) =

2x+ 8a if x < 0

ax2 − 4b if 0 6 x 6 2

−3x+ 18 if x > 2

What values of a and b make f continuouseverywhere?

Hint: because each piece of the function is apolynomial, you know it is continuous on(−∞, 0) ∪ (0, 2) ∪ (2,∞). So, you just need valuesof a and b that will give us

limx→0

f(x) = 0 and limx→2

f(x) = 2.

Khan Academy exercises

Continuity exercises

Continuity of composite functions:

Theorem: If f is continuous at b and

limx→a

g(x) = b, then limx→a

f (g(x)) = f (b).

In other words

limx→a

f (g(x)) = f(limx→a

g(x))

Example: evaluate

limx→1

arcsin(1−√x

1− x

)

Continuity of composite functions:

Theorem: If f is continuous at b and

limx→a

g(x) = b, then limx→a

f (g(x)) = f (b).

In other words

limx→a

f (g(x)) = f(limx→a

g(x))

Example: evaluate

limx→1

arcsin(1−√x

1− x

)

Example: evaluate limx→1

arcsin(1−√x

1− x

).

Solution: We find the limit of the inner function:

limx→1

1−√x

1− x= lim

x→1

1−√x

(1 +√x)(1−

√x)

= limx→1

1

1 +√x=

1

1 +√1=

1

2.

Is f(x) = arcsin x continuous at x = 12? Yes, since

12 ∈ dom f and inverse trig functions are continuouseverywhere in their domains. Therefore

limx→1

arcsin(1−√x

1− x

)= arcsin

(limx→1

1−√x

1− x

)= arcsin

(1

2

)=π

6.

Example: evaluate limx→1

arcsin(1−√x

1− x

).

Solution: We find the limit of the inner function:

limx→1

1−√x

1− x= lim

x→1

1−√x

(1 +√x)(1−

√x)

= limx→1

1

1 +√x=

1

1 +√1=

1

2.

Is f(x) = arcsin x continuous at x = 12? Yes, since

12 ∈ dom f and inverse trig functions are continuouseverywhere in their domains. Therefore

limx→1

arcsin(1−√x

1− x

)= arcsin

(limx→1

1−√x

1− x

)= arcsin

(1

2

)=π

6.

Example: evaluate limx→1

arcsin(1−√x

1− x

).

Solution: We find the limit of the inner function:

limx→1

1−√x

1− x= lim

x→1

1−√x

(1 +√x)(1−

√x)

= limx→1

1

1 +√x=

1

1 +√1=

1

2.

Is f(x) = arcsin x continuous at x = 12?

Yes, since12 ∈ dom f and inverse trig functions are continuouseverywhere in their domains. Therefore

limx→1

arcsin(1−√x

1− x

)= arcsin

(limx→1

1−√x

1− x

)= arcsin

(1

2

)=π

6.

Example: evaluate limx→1

arcsin(1−√x

1− x

).

Solution: We find the limit of the inner function:

limx→1

1−√x

1− x= lim

x→1

1−√x

(1 +√x)(1−

√x)

= limx→1

1

1 +√x=

1

1 +√1=

1

2.

Is f(x) = arcsin x continuous at x = 12? Yes, since

12 ∈ dom f and inverse trig functions are continuouseverywhere in their domains.

Therefore

limx→1

arcsin(1−√x

1− x

)= arcsin

(limx→1

1−√x

1− x

)= arcsin

(1

2

)=π

6.

Example: evaluate limx→1

arcsin(1−√x

1− x

).

Solution: We find the limit of the inner function:

limx→1

1−√x

1− x= lim

x→1

1−√x

(1 +√x)(1−

√x)

= limx→1

1

1 +√x=

1

1 +√1=

1

2.

Is f(x) = arcsin x continuous at x = 12? Yes, since

12 ∈ dom f and inverse trig functions are continuouseverywhere in their domains. Therefore

limx→1

arcsin(1−√x

1− x

)= arcsin

(limx→1

1−√x

1− x

)= arcsin

(1

2

)=π

6.

Continuity of composite functions:

Theorem: If g is continuous at a and f is

continuous at g(a), then the composite

function f ◦ g given by

(f ◦ g)(x) = f (g(x)) is continuous at a.

Examples: where are the following

functions continuous?

h(x) = sin(x2) F (x) = `n(1+cosx)

Continuity of composite functions:

Theorem: If g is continuous at a and f is

continuous at g(a), then the composite

function f ◦ g given by

(f ◦ g)(x) = f (g(x)) is continuous at a.

Examples: where are the following

functions continuous?

h(x) = sin(x2) F (x) = `n(1+cosx)

Continuity of composite functions

F (x) = `n(1 + cos(x))

The Intermediate Value Theorem (informally)

Suppose that you are travelling on a bus fromDurban to Johannesburg along the N3. Let f(x) bea function that gives your elevation (height abovesea level) after distance x from Durban.

This will definitely be a function because it is notpossible to be at two different elevations when youare at single point on the road.

Moreover, this function will be continuous becausef(x) will be defined for all x ∈ [0, 569], and it is notpossible for the bus to jump from one elevation toanother.

The Intermediate Value Theorem (informally)

The red dots are Durban (x = 0, elevation = 0m),Pietermaritzburg, Mooi River, Van Reenen,Harrismith, Warden, Villiers and Johannesburg(x = 569 and elevation=1751m).

The Intermediate Value Theorem (informally)

The (continuous) bus journey starts at an elevationof 0m and finishes at an elevation of 1751m. Whatthe Intermediate Value Theorem says is that if youtake any elevation N between 0 and 1751, theremust be some x-value (i.e. some distance fromDurban along the N3) where your elevation will beequal to N .

The Intermediate Value Theorem:

Suppose that f is continuous on the

closed interval [a, b] and let N be any

number between f (a) and f (b), where

f (a) 6= f (b). Then there exists a number

c ∈ (a, b) such that f (c) = N .

The Intermediate Value Theorem:

Notice that in the second picture (b) there

are three x-values (c1, c2 and c3) where we

have the y-value of the function equal to N .

The Intermediate Value Theorem:

Notice that in the second picture (b) there

are three x-values (c1, c2 and c3) where we

have the y-value of the function equal to N .

Notice that in the second picture (b) there are threex-values (c1, c2 and c3) where we have the y-valueof the function equal to N .

The theorem states “. . . there exists c ∈ (a, b) suchthat f(c) = N”. It is saying that there must be atleast one x-value for which f(x) = N , so havingmore than one still allowed.

Suppose that f is continuous on the closedinterval [a, b] and let N be any number betweenf(a) and f(b), where f(a) 6= f(b). Then thereexists a number c ∈ (a, b) such that f(c) = N .

NB: the conditions of IVT require f(a) 6= f(b).Therefore it can also be used when f(a) > f(b).

Example: Show that there is a root of the

equation

4x3 − 6x2 + 3x− 2 = 0

in the interval (1, 2).

How do we approach this question?

Example: Show that there is a root of the

equation

4x3 − 6x2 + 3x− 2 = 0

in the interval (1, 2).

How do we approach this question?

The question is asking us to show that there

exists a value of x in the interval (1, 2) such

that f (x) = 0.

We must see if we can apply the

Intermediate Value Theorem!

If it works, we will be applying the IVT with

a = 1, b = 2, x = c and 0 = N .

Before we can apply the IVT, we must check

to see if these parameters satisfy the

conditions that are required by the IVT.

We must check the following:

I f (x) = 4x3 − 6x2 + 3x− 2 is continuous

on the closed interval [1, 2]

I f (1) 6= f (2)

I 0 (the value for N) is between f (1) and

f (2)

Once we know that those three conditions

have been satisfied, we can apply the

Intermediate Value Theorem.

I The function f(x) = 4x3 − 6x2 + 3x− 2 iscontinuous on the closed interval [1, 2] becauseit is a polynomial and therefore is continuouseverywhere.

I We calculatef(1) = 4(1)− 6(1) + 3(1)− 2 = −1 andf(2) = 4(8)− 6(4) + 3(2)− 2 = 12. Clearlyf(1) 6= f(2).

I Our value of N is 0. It is clear that−1 < 0 < 12.

We have shown that the function f(x) satisfies theconditions of the Intermediate Value Theorem (witha = 1, b = 2 and N = 0) and therefore there mustexist c ∈ (1, 2) such that f(c) = 0.

I The function f(x) = 4x3 − 6x2 + 3x− 2 iscontinuous on the closed interval [1, 2] becauseit is a polynomial and therefore is continuouseverywhere.

I We calculatef(1) = 4(1)− 6(1) + 3(1)− 2 = −1 andf(2) = 4(8)− 6(4) + 3(2)− 2 = 12. Clearlyf(1) 6= f(2).

I Our value of N is 0. It is clear that−1 < 0 < 12.

We have shown that the function f(x) satisfies theconditions of the Intermediate Value Theorem (witha = 1, b = 2 and N = 0) and therefore there mustexist c ∈ (1, 2) such that f(c) = 0.

I The function f(x) = 4x3 − 6x2 + 3x− 2 iscontinuous on the closed interval [1, 2] becauseit is a polynomial and therefore is continuouseverywhere.

I We calculatef(1) = 4(1)− 6(1) + 3(1)− 2 = −1 andf(2) = 4(8)− 6(4) + 3(2)− 2 = 12. Clearlyf(1) 6= f(2).

I Our value of N is 0. It is clear that−1 < 0 < 12.

We have shown that the function f(x) satisfies theconditions of the Intermediate Value Theorem (witha = 1, b = 2 and N = 0) and therefore there mustexist c ∈ (1, 2) such that f(c) = 0.

I The function f(x) = 4x3 − 6x2 + 3x− 2 iscontinuous on the closed interval [1, 2] becauseit is a polynomial and therefore is continuouseverywhere.

I We calculatef(1) = 4(1)− 6(1) + 3(1)− 2 = −1 andf(2) = 4(8)− 6(4) + 3(2)− 2 = 12. Clearlyf(1) 6= f(2).

I Our value of N is 0. It is clear that−1 < 0 < 12.

We have shown that the function f(x) satisfies theconditions of the Intermediate Value Theorem (witha = 1, b = 2 and N = 0) and therefore there mustexist c ∈ (1, 2) such that f(c) = 0.

Please note: we can’t claim that a function

is continuous on [a, b], without saying why it

is continuous there.

To do this you just need

to use the types of functions from which the

function is made. In the last example it is

simple: f (x) is a polynomial and hence

continuous everywhere.

For a function like h(x) = `n(x) + 2x− 3 it

is not so simple. Why is

`n(x) + 2x− 3 continuous on (0,∞)?

Please note: we can’t claim that a function

is continuous on [a, b], without saying why it

is continuous there. To do this you just need

to use the types of functions from which the

function is made. In the last example it is

simple: f (x) is a polynomial and hence

continuous everywhere.

For a function like h(x) = `n(x) + 2x− 3 it

is not so simple. Why is

`n(x) + 2x− 3 continuous on (0,∞)?

Please note: we can’t claim that a function

is continuous on [a, b], without saying why it

is continuous there. To do this you just need

to use the types of functions from which the

function is made. In the last example it is

simple: f (x) is a polynomial and hence

continuous everywhere.

For a function like h(x) = `n(x) + 2x− 3 it

is not so simple.

Why is

`n(x) + 2x− 3 continuous on (0,∞)?

Please note: we can’t claim that a function

is continuous on [a, b], without saying why it

is continuous there. To do this you just need

to use the types of functions from which the

function is made. In the last example it is

simple: f (x) is a polynomial and hence

continuous everywhere.

For a function like h(x) = `n(x) + 2x− 3 it

is not so simple. Why is

`n(x) + 2x− 3 continuous on (0,∞)?

Why is `n(x) + 2x− 3 continuous on (0,∞)?

I f(x) = `n(x) is a logarithmic function andtherefore is continuous everywhere on itsdomain. Therefore it is continuous on (0,∞).

I g(x) = 2x− 3 is a polynomial and therefore iscontinuous everywhere, and in particular on(0,∞).

I The theorem about building continuousfunctions says that if f and g are bothcontinuous at a, then f + g is also continuousat a. Therefore h(x) = f(x) + g(x) will becontinuous on (0,∞).

Why is `n(x) + 2x− 3 continuous on (0,∞)?

I f(x) = `n(x) is a logarithmic function andtherefore is continuous everywhere on itsdomain. Therefore it is continuous on (0,∞).

I g(x) = 2x− 3 is a polynomial and therefore iscontinuous everywhere, and in particular on(0,∞).

I The theorem about building continuousfunctions says that if f and g are bothcontinuous at a, then f + g is also continuousat a. Therefore h(x) = f(x) + g(x) will becontinuous on (0,∞).

Why is `n(x) + 2x− 3 continuous on (0,∞)?

I f(x) = `n(x) is a logarithmic function andtherefore is continuous everywhere on itsdomain. Therefore it is continuous on (0,∞).

I g(x) = 2x− 3 is a polynomial and therefore iscontinuous everywhere, and in particular on(0,∞).

I The theorem about building continuousfunctions says that if f and g are bothcontinuous at a, then f + g is also continuousat a. Therefore h(x) = f(x) + g(x) will becontinuous on (0,∞).

Why is `n(x) + 2x− 3 continuous on (0,∞)?

I f(x) = `n(x) is a logarithmic function andtherefore is continuous everywhere on itsdomain. Therefore it is continuous on (0,∞).

I g(x) = 2x− 3 is a polynomial and therefore iscontinuous everywhere, and in particular on(0,∞).

I The theorem about building continuousfunctions says that if f and g are bothcontinuous at a, then f + g is also continuousat a. Therefore h(x) = f(x) + g(x) will becontinuous on (0,∞).

Important points about applying IVT:

I You don’t have to have f(a) < f(b), all youneed is f(a) 6= f(b)

I Often you must manipulate the equation to getit as f(x) = p where p ∈ R. Note: p doesn’thave to be zero (see Ex. 51).

I Sometimes you won’t be able to calculate theexact value of f(a) or f(b) (see Exercise 51,54, 56–58). This is not a problem: you onlyneed to show that f(a) 6= f(b) and that N isbetween f(a) and f(b). If N = 0 then you canjust show that that one of f(a) or f(b) ispositive and the other one is negative.