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194

CHAPTER 3
...
3

Continuity on Intervals
“A function φ(x) is said to be continuous between any limiting values of x, such as a
and b, when to each value of x between those limits there corresponds a finite value of
the function, and when an indefinitely small change in the value of x produces only
an indefinitely small change in the function
...
This continuity can be readily illustrated by taking φ(x)
as the ordinate of a curve, whose equation may then be written y = φ(x)
...


3
...
1

Continuity on Intervals: Main Definitions and Theorems

The significance of continuity is perhaps best understood when applied to whole intervals (a, b),
or [a, b], etc
...
Below we will define what it means for f (x) to be
continuous on (a, b), and continuous on [a, b]
...
) Continuity on open
intervals is rather trivial to define, but nonetheless has interesting consequences
...
But
first we look at the open intervals
...
3
...
14
What is interesting about continuity on (a, b) is it implies f ((a, b)) = is an interval of some kind
(possibly infinite, or even a single point), and that the curve y = f (x) is a connected graph for
a < x < b
...
3
...

Unfortunately the proof is a few steps beyond the scope of this textbook and is left, for the
interested reader, to a course in advanced calculus, real analysis or topology
...
Put another way, we could draw the graph without lifting our pen from the paper
...
when an indefinitely small change in the value of x produces only an indefinitely small change
14 Of course each x may require a different δ for a given ε in the original definition of continuity given in
0
Section 3
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To summarize, (∀x0 ∈ (a, b))(∀ε > 0)(∃δ > 0)(∀x)(|x − x0 | < δ −→ |f (x) − f (x0 )| < ε)
...
In other words,
f (S) is the set of all possible outputs of f ( ) if the inputs are taken from S
...
Similarly, if f (x) = sin x then f (R) = [−1, 1]
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3
...
CONTINUITY ON INTERVALS

a
b
f ((a, b)) a finite,
open interval

195

a
b
f ((a, b)) a finite,
closed interval
...


Figure 3
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The image will always be an interval of some kind—finite,
infinite, open, closed, half-open or a “single-point” interval [c, c] if f is a constant function
...
As a consequence it is
then reasonable that the range will be a (connected) interval as the theorem also states
...
2 shows sample cases for continuity on open intervals, illustrating that the image is
always an interval
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3
...
f (x) is continuous on the open interval (a, b), i
...
, continuous at each x ∈ (a, b), and
2
...
f (x) is left-continuous at x = b
...
e
...
In this way we ignore the behavior of f ( ) outside of inputs from strictly
from [a, b]
...
Similarly for continuity on [a, b)
...
Both are wrapped
up nicely in the following analog to Theorem 3
...
1, the difference being that the previous theorem required continuity on (a, b), while this theorem requires continuity on [a, b]
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3
...

In other words, the continuous image of a closed and bounded interval [a, b] is also a closed
and bounded interval [c, d]
...
CONTINUITY AND LIMITS OF FUNCTIONS
f (xmax )
f (b)

f (a)
f (xmin )
a

xmin

xmax

b

Figure 3
...

The image f ([a, b]) is also a finite, closed interval (Theorem 3
...
2), containing a maximum
value f (xmax ) and a minimum value f (xmin ), which are achieved at some xmax , xmin ∈ [a, b]
(Extreme Value Theorem, Corollary 3
...
1)
...
3
...


this is also believable
...
In doing so we would
somewhere draw the highest and lowest points of our curve, and these heights would be d and c,
respectively, from our theorem
...

The proof of Theorem 3
...
2 is also beyond the scope of this text, but the EVT and IVT
follow very quickly from the theorem
...

Corollary 3
...
1 (Extreme Value Theorem) If f : [a, b] −→ R is continuous on [a, b], then
f (x) achieves its maximum and minimum heights at some xmax , xmin ∈ [a, b]
...


Corollary 3
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2 (Intermediate Value Theorem) If f (x) is continuous on [a, b], then f (x)
achieves every value between f (a) and f (b)
...

(Note that the statement of IVT is contained in the second sentence of the quote from Britannica
on page 194, and the EVT is also hinted at in the quote
...
3
...
e
...
The closed criterion is clearly necessary as shown in the the first and third
graphs in Figure 3
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In each graph in that figure, a function f (x) defined on an open
interval (a, b) is graphed: in the first graph, the image is an open interval and so no maximum

3
...
CONTINUITY ON INTERVALS

197

2
1

−1

2
1

1

2

−1

3

−1

−2

1

2

3

−1
−2

f (x) =

1
x−1

x ∈ [−1, 3] − {1}

f (x) =

x2 , −1 ≤ x < 1
2 + 0
...
5

Figure 3
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3
...
3
...
In the first
graph, there is a vertical asymptote at x = 1, which is a point of discontinuity
...
In the second case, there is a “jump” discontinuity at x = 1, and the image
[0, 1] ∪ [2, 2
...
It happens that the
second graph does have maximum and minimum values f (0) = 0 and f (3
...
5 though this
was not guaranteed because f (x) is not continuous on all of [−1, 3
...
These examples do not
violate the corollaries IVT and EVT since both corollaries claim the truth of tautologies of the
form P → Q, which is equivalent to (∼ P ) ∨ Q, and here we have ∼ P in both cases, making
them true to the theorems and corollaries vacuously (in form P → Q) or trivially (using the
form (∼ P ) ∨ Q)
...
It may happen that f ((a, b)) is a
closed and bounded interval, as in the second graph in that figure, but it clearly (from the other
two graphs in that figure) is not guaranteed
...
4
...
Similarly it is continuous on (1, 3]
...
5]
...
5], but rather
that the “piece” drawn on that interval is a continuous “piece,” in the sense of Definition 3
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2,
page 195 and the discussion following that
...
3
...
Here we will
instead look at the IVP and its usefulness in algebra
...
3
...


198

CHAPTER 3
...
The proof is a matter of chasing down the
definitions of continuity on the various intervals, and checking each of the cases
...

Example 3
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1 Show that the equation x5 + 7 = x2 has a solution in R
...

Next define f (x) = x5 − x2 + 7, which is a polynomial and thus continuous on all of R (think of
R as the open interval (−∞, ∞))
...
Next notice that
f (−2) = (−2)5 − (−2)2 + 7 = −29,

and

f (1) = 1 − 1 + 7 = 7
...
Since f (−2) = −29 while
f (1) = 7, there exists some x0 between −2 and 1 such that f (x0 ) = 0 (by IVT), and this x0
therefore solves the original equation, q
...
d
...
3
...
It is similar to the usual algebraic trick where we
try to solve an equation of the form g(x) = h(x) by instead determining where g(x) − h(x) = 0
...
As we saw in the algebra
sections of Chapter 2, solving f (x) = 0 is usually simpler than finding where g(x) = h(x)
...
3
...

Solution: This is equivalent to the equation x3 − 8x2 + 15x + 1 = 0 having at least three
solutions
...

First we notice that f (x) is a polynomial, and therefore continuous on R
...


3
...
3

We see that f (x) must be zero for some
x ∈ (−1, 0), another x ∈ (0, 4), and yet another x ∈ (4, 5), ultimately therefore proving
that there must be at least three solutions of
x3 − 8x2 + 15x = −1, q
...
d
...
We can exploit this to solve polynomial and rational inequalities
...

Example 3
...
3 Solve the inequality x3 − 9x2 + 20x ≤ 0
...
3
...
We wish to see where
f (x) ≤ 0
...


3

2

⇐⇒ x − 9x + 20x = 0

⇐⇒ x(x2 − 9x + 20) = 0

⇐⇒ x(x − 4)(x − 5) = 0
⇐⇒
x = 0, 4, 5
...
We can use this fact and a simple test, for example, to find the
sign of f (x) on all of (−∞, 0): if we know the sign of f (x) at, say, x = −10, then we know it
for the whole interval x ∈ (−∞, 0) because whatever is the sign at x = −10 will be the sign for
that whole interval (because if it had two signs in that interval then continuity would guarantee
there would be another point where f (x) = 0 in that interval, and we know that the only points
where that happens, x = 0, 4, 5 are not in that interval)
...

A nice visual device for determining where f (x) is positive, and where it is negative, is a sign
chart, as illustrated below
...
In the style of sign chart is given below,
we use ⊕ to represent a positive quantity, and ⊖ to represent a negative quantity
...
5
⊕⊖⊖
⊕⊕⊖

−10
⊖⊖⊖
⊖

0

⊕

4

⊖

10
⊕⊕⊕
5

⊕

From the chart we see that f (x) < 0 on the first and third intervals, i
...
, for x ∈ (−∞, 0) ∪ (4, 5)
...

The information at hand will not give us a complete picture of the graph of f (x), but it
is instructive to see what the graph looks like, and how an accurate enough (for our purposes
here) picture can be easily imagined from the sign chart
...
5
...
Since f is continuous,
the only way it can change sign is to pass through zero (by IVT, see also Figure 3
...
These mark boundaries of subintervals of R on which f does
not change sign
...
If f (x) happens
to be factored, we only need to check the signs of each factor to see if the negative factors
“cancel” completely to leave a positive function, or if we have an odd number of negative factors
to make f negative on the interval in question
...
CONTINUITY AND LIMITS OF FUNCTIONS

15
10
5

−1

1

2

3

4

5

−5

−10
−15

Figure 3
...
Compare to the sign chart in Example 3
...
3
...
3
...
There can be complications, and we have to
be careful to answer the given question
...

Example 3
...
4 Solve x2 ≥ x + 1
...

Now solving f (x) = 0 requires the quadratic formula or completing the square
...
Recall first that f (x) = 0 ⇐⇒ x2 − x − 1 = 0
...
61803, 1
...

2(1)
2 2

We will always use the exact values, but the approximate ones are also useful since we need to
know where to find our test points
...

2

Such an approach is perhaps more sophisticated than our method in Example 3
...
4, where we did not bother to
factor f (x), but is often unwieldy and requires more subtlety than necessary to solve the inequality
...
3
...
6: Actual graph of f (x) = x2 − x − 1
...
61803, 1
...
Compare to the sign chart in Example 3
...
4
...
These include the cases where
f (x) = 0 as well as where f (x) > 0
...


The function f (x) = x − x − 1 is graphed in Figure 3
...
Compare the graph with
the sign chart above
...

This is not always the case
...
Consider the following example
...
3
...

Solution: This is already a question about sign, so we will simply take f (x) = x3 − 6x2 + 9x
and solve f (x) > 0
...

We see that f (x) = 0 at x = 0, 3
...


4
⊕⊕2
⊕

202

CHAPTER 3
...
7: Graph of f (x) = x(x − 3)2 as in Example 3
...
5, illustrating that a function’s
height can be zero without the function changing signs there
...
3
...
7
...
Some terminology helps to express this
...
3
...

Then we can observe the following for a polynomial f (x):
1
...
if x = a is a zero of even multiplicity, then f (x) does not change sign at x = a
...

In Example 3
...
5 we had x = 3 was a zero of multiplicity 2, which is even and so f (x) did not
change sign (in the sense of changing from positive to negative or vice versa) as x passed through
the value 3, while x = 0 was a zero of multiplicity 1, which is odd and so the function did change
sign there
...

It is sometimes the case that, even when f (x) is factored, some of the factors might never
be zero
...
One common such type of
factor is of the form x2k + a, where k ∈ N and a > 0
...
Another type is a factor of the form ax2 + bx + c where b2 − 4ac < 0, in which case

3
...
CONTINUITY ON INTERVALS

203

ax2 + bx + c = 0 has no real solutions
...
The following examples illustrates
one case
...
3
...

Solution: As before, we construct f (x) = x5 − 25x, so that
x5 ≤ 25x ⇐⇒ x5 − 25x ≤ 0 ⇐⇒ f (x) ≤ 0
...
Now the first factor is zero for x = 0,
√
the second factor is never zero, and the third is zero for x = ± 5 ≈ ±2
...
Hence we get
the following sign chart:
Function:
Test x =
Sign Factors:
Sign f (x):

f (x) = x(x2 + 5)(x2 − 5)
−1
1
⊖⊕⊖
⊕⊕⊖

−4
⊖⊕⊕
⊖

√
− 5

We see that f (x) ≤ 0 for x ∈ ∞, −

3
...
4

√

⊕

0

⊖

4
⊕⊕⊕
√
5

⊕

√
5 ∪ 0, 5
...
For instance,
if we define f (x) = 1/(x − 1), we see that f (−1) = −1/2, while f (3) = 1/2
...
The reason the IVT did not apply is
that f (x) = 1/(x − 1) is not continuous in [−1, 3], since it is discontinuous at x = 1
...
e
...
See the first graph in Figure 3
...
(The second graph also shows how
discontinuity allows “jumping
...
We simply need to analyze them
and the IVT further
...
3
...
Then at least one of the following must hold:
(i) there exists c between a and b so that f (c) = C; or
(ii) f (x) has at least one discontinuity on [a, b]
...
e
...


“
√ ”
√ ”“
this example we could continue factoring f (x) = x(x2 + 5)(x2 − 5) = x(x2 + 5) x − 5 x + 5
...
It is really a matter
of personal taste
...
CONTINUITY AND LIMITS OF FUNCTIONS
Proof: We can use some symbolic logic to prove this
...

Thus f continuous on [a, b] =⇒ (i), meaning that f continuous on [a, b] −→ (i) is a
tautology
...
e
...


This proof is, of course, true and precise, but the corollary should also be intuitive: as we
move x along the interval [a, b], to pass continuously from one height to another we pass through
all intermediate heights; if we are not going to pass through all intermediate heights, we have to
somehow “jump” over any missed heights, requiring a discontinuity
...
For a function’s output
to change signs as we vary its input, that output must pass through zero or be discontinuous
...
For rational functions this means we look at where the
numerator is zero, and where the denominator is zero, respectively
...
3
...

x −1
x
x
=
, we see that f is zero at x = 0, and
Solution: Defining f (x) = 2
x −1
(x + 1)(x − 1)
discontinuous at x = ±1
...

f (x) =

Function:

x
(x + 1)(x − 1)

Test x =

−2

−0
...
5

2

Sign Factors:

⊖
⊖⊖

⊖
⊕⊖

⊕
⊕⊖

⊕
⊕⊕

Sign f (x):

⊖

−1

⊕

0

⊖

1

⊕

Since we are interested in the points where f (x) ≤ 0, we need the open intervals on which
f (x) < 0, i
...
, the first and third intervals, and all points where f (x) = 0, i
...
, x = 0
...

x
We include a graph of f (x) = (x+1)(x−1) in Figure 3
...
There are other types of discontinuities besides vertical
asymptotes, but for rational functions f in which the numerator and denominator have no
common factors, vertical asymptotes are the only type of discontinuity which can occur
...

For another perspective justifying the technique for sign charts for rational functions, consider
that a ratio of functions can only change signs if the numerator or denominator changes signs
...
Summarizing, we conclude that a ratio of polynomials can
only change signs if the numerator passes through zero or the denominator passes through zero
...

The method above can generalize for solving any rational inequality the same way we generalized for the polynomial case: we rewrite any inequality into an equivalent statement about
signs (+/−)
...
3
...
See Example 3
...
7
...
Vertical
asymptotes will be properly developed later in the text
...
8: Graph of f (x) =

x
2x
≤ 2

...
Hence we have

Example 3
...
8 Solve the inequality

x2

x
2x
x(x2 − 9) − 2x(x2 − 7)
−x3 + 5x
−x(x2 − 5)
− 2
=
= 2
= 2
,
x2 − 7 x − 9
(x2 − 7)(x2 − 9)
(x − 7)(x2 − 9)
(x − 7)(x2 − 9)
√
and we are trying to find where f (x) ≤ 0
...
23607, and is discontinuous at x = ±3 and x = ± 7 ≈ ±2
...
The sign chart follows:
f (x) =

f (x) =

Function:
Test x =
Sign Factors:

Sign f (x):

−2
...
5
⊕⊕
⊖⊖
√
− 7

⊕

−1
⊕⊖
⊖⊖
√
− 5

1
⊖⊖
⊖⊖
√
5

0
⊖

2
...
9
⊖⊕
⊕⊖
√
7

⊖

10
⊖⊕
⊕⊕
3

⊕

⊖

It is important to note that the sign chart gives us the signs on the various open intervals
...
Since for this case we want our solution to be those points where f (x) ≤ 0, we
have to include those endpoints where f (x) = 0
...


√
Note where f changes signs continuously (i
...
, passing √
through zero height) at 0, ± 5 and
discontinuously (in fact, via vertical asymptotes) at x = ± 7, ±3
...
CONTINUITY AND LIMITS OF FUNCTIONS

f (x) here, but much of its behavior is evident by the sign chart and the way f changes signs by
√
passing through height zero at x = 0, ± 5, and by discontinuity (in fact, via vertical asymptotes)
√
at x = ±3, ± 7
...
For completeness here we also discussed the inclusive inequalities (≥, ≤)
...

x2 + x + 1
> 0
...
For
the numerator (NUM) we need the quadratic formula:
Example 3
...
9 Solve the inequality

NUM = 0 ⇐⇒ x =

−1 ±

√
12 − 4(1)(1)
1 − −3
=
∈ R
...
For the denominator (DEN) we have
DEN = 0 ⇐⇒ x2 − 7x + 12 = 0 ⇐⇒ (x − 3)(x − 4) = 0 ⇐⇒ (x = 3) ∨ (x = 4)
...
5
⊕/(⊕⊖)
3

⊕

5
⊕/(⊕⊕)
4

⊖

⊕

From the sign chart we see the solution is x ∈ (−∞, 3) ∪ (4, ∞)
...
For each of the following, draw a continuous function f (x) whose domain is
x ∈ (2, 5), and whose image of that set
(i
...
, whose range) is given
...
Repeat the previous problem except
assume the domain for f (x) is x ∈
[2, 5]
...
See Theorem 3
...
2, page 195
...
Draw the graph of a function y = f (x),
defined for x ∈ [2, 5], where f (x) is continuous on (2, 5) but not on [2, 5]
...
Show that x3 − 6x2 + 4x+ 10 = 0 has at
least three solutions, by checking values

3
...
CONTINUITY ON INTERVALS

207

of f (x) = x3 − 6x2 + 4x + 10 at various
x-values
...


27x − x2
<0
x2 + 11x + 30

5
...


18
...


1
x
>
x+5
x−7

For each of the following, solve the inequality
by means of a sign chart
...

6
...
x3 + 2x2 ≤ 15x
21
...
x2 − 9 < 0
8
...
x2 + 3x ≥ 2
2

10
...
x2 + 15 > 0
12
...

x+5
x−7
14
...


x2 − 16
>0
x2 + 16

16
...


x4

x2 − 1
≤0
− 3x2 − 10

3x
2x
<
x+5
x+6

Use a sign chart to graph the following functions, to the extent that continuity and sign
of f are illustrated
...

23
...
f (x) = x3 − x2 − 2x
25
...
f (x) =

x
x2 − 9

x−1
x2 − 9x + 8



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