Transcript x - mor media international

Intervals of Continuity
Page 8
Even though not all functions are continuous on all real numbers (i.e., everywhere), we can still talk about
the continuity of a function in terms of intervals. As you may recall, there are three kinds of intervals:
open intervals, closed intervals, half-open (or half-closed) intervals.
Definition of f(x) is continuous on an open interval:
We say a function f(x) is continuous on an open interval of (a, b) if there are no points
of discontinuity on the interval (a, b). That is, f(x) is continuous at any x where a < x < b.
a
b
a
b
a
b
a
b
We say a function f(x) is continuous on a closed interval of [a, b] if
i. f(x) is continuous on the open interval (a, b),
ii. f(a) and f(b) are defined, and
a
b
iii. limxa+ f(x) = f(a) and limxb– f(x) = f(b).
We say a function f(x) is continuous on a half-open interval of [a, b) if
i. f(x) is continuous on the open interval (a, b), and
ii. f(a) is defined and limxa+ f(x) = f(a).
a
b
We say a function f(x) is continuous on a half-open interval of (a, b] if
i. f(x) is continuous on the open interval (a, b), and
ii. ______________________________.
a
b
Continuous on its Domain
Recall that functions such as f(x) = x1/2 and f(x) = log x are not continuous
on all real numbers, nevertheless, they are continuous at every number in
their domains. For example, the domain of f(x) = x1/2 is [0, ) and its
interval of continuity is also [0, ). Similarly, the domain of f(x) = log x is
(0, ) and its interval of continuity is also (0, ). Another example is f(x)
= 1/x, where its graph has two pieces and there is a vertical asymptote at x
= 0. However, since its domain is all real numbers except 0, i.e., (, 0) 
(0, ), and we can see that the left piece of f(x) = 1/x is continuous on
(, 0) where as the right piece is continuous on (0, ), we still say that
the function is continuous on its domain.
So what is an example of a function which is not continuous on its
domain?
There are many functions which are not continuous on their domains. For
example, the integer function, f(x) = [x]. The domain is all real numbers
(plug any real number x into the integer function, it will yield back the
greatest integer ≤ x). However, the function is not continuous on its
domain since it is discontinuous at every integer. Another example is the
sign function,
f(x) = sgn x =
1, x  0

0, x  0
 1, x  0

where given any real number x, this function will
yield back 1 if x is positive, 1 if x is negative, and 0 is x is zero. As you
can see, its domain is all real numbers, but it’s not continuous at x = 0.
Page 9
How to Give the Intervals of Continuity of a Function
Page 10
If a function is continuous on all real numbers, e.g., f(x) = x2, then we say its interval of continuity (IOC)
is (, ).
If a function is continuous on its domain, then its interval(s) of continuity is same as its domain. For
example, i) f(x) = x1/2  IOC = [0, ); ii) f(x) = 1/x  IOC = (, 0)  (0, ).
If a function is not continuous on its domain, e.g., f(x) = sgn x, we must say the IOC is (, 0)  (0, )
despite that the domain is (, ). This implies the interval notation for the domain is not necessarily
same as the intervals of continuity.
Example
For the function with graph below, give the domain and intervals of continuity using the least number of
intervals as possible.
Domain: ____________________________
IOC: _______________________________
3
y = f(x)
2
1
–6 –5 –4 –3 –2 –1 0
–1
1
2
3
4
5
6
7
8
9
–2
–3
Domain: (-oo, -1) U (-1,5) U (5, oo)
IOC: (-oo, -4] U (-4,-1) U (-1, 1)U (1, 3) U (3, 5) U (5, 8) U [8.oo)
10
11
Properties of Functions which are Continuous on their Domains
Page 11
Theorem:
The following types of functions are continuous at every number in their domains:
i) polynomial functions
ii) rational functions
iii) root functions
iv) trigonometric functions v) exponential functions
vi) logarithmic functions
vii) absolute-value functions viii) inverse trigonometric functions
Properties:
If f(x) and g(x) are two functions continuous at every
number in their respective domains, and let D be the
domain of f(x) + g(x) and E be the domain of f(g(x)), then:
i)f(x) + g(x), f(x) – g(x) and f(x)∙g(x) are continuous on D,
ii)f(x)/g(x) is continuous on D – {x | g(x) = 0}, and
iii)f(g(x)) is continuous on E.
Continuous on all
real numbers
Continuous only on
their domains
Polynomial
Rational
Odd-indexed root
Even-indexed root
sin, cos
tan, cot, sec, csc
tan–1, cot–1
sin–1, cos–1, sec–1, csc–1
Exponential
Logarithmic
Absolute-value
Example
Let f(x) = sin x and g(x) = x2 – 1 . Find the intervals of continuity of the following functions:
1. f(x) + g(x)
2. f(x) – g(x)
3. f(x)∙g(x)
4. f(g(x))
5. g(f(x))
6. f(x)/g(x)
7. g(x)/f(x)
How Do We Find the Limit of a Function Without the Graph?
Page 12
There are several ways we can find the limit of a function f as x approaches a without knowing
its graph. One obvious way is to use numbers close to a. See the following examples:
1. limx1 (x2 + 1) = ___
x
0.9
0.99
0.999
2. limx2 1/x2 = ___


1

x2 + 1
1.001
1.01
x
1.1
2.9
2.99
2.999


3

–1.1
–1.01
3.001
3.01
3.1
5
x 1
–1

7
x2
1.9
1.99
1.999



2
2.001
2.01
2.1

1.9
1.99
1.999
1
x 4


–.999
–.99
–.9
–.1
x
–.001
x
8
x2
2


2.001
2.01

2.001
2.01
2.1
(where x in radians)

0

x
3x
2x  9
100
2.1


0.001
0.01

9. limx∞ 3x/(2x – 9) = ___
8. limx2+ 8/(x + 2) = ___
2
–.01
6 sin x
x

2

2
6. limx0 6 sin x/x = ___
7. limx2– 7/(x – 2) = ___
x


x


–1.001
1.999
4. limx2 1/(x2 – 4) = ___
5. limx–1 5/(x + 1) = ___
x
1.99
1/x2

3. limx3 (x – 3)/|x – 3| = ___
x
x3
| x 3|
1.9
1,000
10,000


∞
0.1
Implication of Continuity on Limits
Page 13
Recall that if f(x) is continuous at x = a, then the limit of f(x) as x approach a must exist and it must equal
to f(a). Therefore, when we need to evaluate limxa f(x) and if we know f(x) is continuous at x = a, all we
need to do is to evaluate f(a), i.e., whatever f(a) is, is the limit!
Direct Substitution Property:
Rule of Thumb of Evaluating Limit When you evaluate
If f(x) is continuous at x = a,
then limxa f(x) = f(a).
the limit of a function f(x) as x approaches a, just plug in
a into the f(x) first. That is, when limxa f(x) is asked, just
do f(a).
With this property, it allows us to
evaluate the limit by using the socalled “plug-in” method. That is, as
long as we know f(x) is continuous
at a, even if we don’t know the
graph of f(x), we can just evaluate
f(a) and that will be limit! We don’t
need to use any numbers close to a,
hence, it is a much better and faster
way of finding the limit than the
“tabular” method used on page 5.
Examples:
1. limx1 (x2 + 1) = 2
2. limx2 1/x2 = 1/22 = ¼
3. limx3+ (x2 – 2)/(x + 1) = 7/4
4. limx– cos x =
The rule of thumb above really is another way (actually, my way)
to state the Direct Substitution Property. The difference is: here we
don’t even need to care whether f(x) is continuous at a or not.
By Direction Substitution Property
By My Rule of Thumb
Example 1: x2 + 1 is a quadratic
function. Quadratic functions are a type
of polynomial functions and polynomial
functions are continuous everywhere,
therefore it’s okay to plug the 1 into x,
and obtain 12 + 1 = 2 as the limit.
Example 2: 1/x2 is a rational function
with a vertical asymptote at x = 0. So
1/x2 is discontinuous at x = 0 but it’s
continuous elsewhere, therefore it’s
okay to plug the 2 into x, and obtain
1/22 = ¼ as the limit.
Example 1: 12 + 1 = 2
Example 2: 1/22 = 4