Name: _________________________________________________ Period: ___________ Date: ________________
Continuity, End Behavior, and Limits Guided Notes
The graph of a continuous function has no breaks, holes, or gaps. You can trace the graph of a continuous function
without lifting your pencil.
One condition for a function π(π) to be continuous at π = π is that the function must approach a unique function value
as π -values approach π from the left and right sides. The concept of approaching a value without necessarily ever
reaching it is called a limit.
If the value of π(π)approaches a unique value π³ as π approaches π from each side, then the limit of π(π) as π
approaches π is π³. π₯π’π¦ π(π) = π³
π→π
Functions that are not continuous are discontinuous. Graphs that are discontinuous can exhibit:
ο·
Infinite discontinuity (A function has an infinite discontinuity at π = π if the function value increases or
decreases indefinitely as π approaches π from the left and right)
ο·
Jump discontinuity,( A function has a jump discontinuity at π = π if the limits of the function as π approaches π
from the left and right exist but have two distinct values.
ο·
Removable discontinuity, also called point discontinuity (function has a removable discontinuity if the function is
continuous everywhere except for a hole at π = π.
Continuity Test
A function π(π) is continuous at π = π if it satisfies the following conditions.
1. π(π) is defined at π. π(π)exists.
2. π(π) approaches the same function value to the left and right of π.
π₯π’π¦ π(π) ππππππ
3. The function value that π(π)approaches from each side of π is π(π).
π₯π’π¦ π(π) = π (π)
π→π
π→π
Sample Problem 1: Determine whether each function is continuous at the given π -values. Justify using the continuity
test. If discontinuous, identify the type of discontinuity.
a.
π(π) = πππ + π − π ππ π = π
π(π) = πππ + π − π
ππ π = π
π
π(π) = π ∗ π + π − π
π(π) = −π
π(π) ππππππ
π → π− π → −π
y
30
20
10
x
-40
-30
-20
-10
10
20
30
π
π. π
π. ππ
π. πππ
π(π)
−π. ππ
−π. ππππ
−π. ππππππ
π
π. π
π. ππ
π. πππ
π(π)
−π. ππ
−π. ππππ
−π. ππππππ
π → π+ π → −π
40
-10
-20
-30
π(π) = −π πππ
π → −π ππππ ππππ πππ
π ππ π = π
π₯π’π¦ πππ + π − π = π (π)
π→π
π(π) = πππ + π − π ππ ππππππππππ ππ π = π
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Name: _________________________________________________ Period: ___________ Date: ________________
Continuity, End Behavior, and Limits Guided Notes
b.
π(π) =
|ππ|
π
|ππ|
ππ π = π
π
|π ∗ π| π
π(π) =
=
π
π
π»ππ ππππππππ ππ πππ
ππππππ
ππ π = π
π → π− π → −π
ππ π = π
5
π(π) =
y
4
3
2
1
x
-5
-4
-3
-2
-1
1
2
3
4
5
π
−π. π
−π. ππ
−π. πππ
π(π)
−π
−π
−π
π
π. π
π. ππ
π. πππ
π(π)
π
π
π
π → π+ π → π
-1
-2
-3
-4
|ππ|
πππ ππππ π
ππππππππππππ ππ π = π
π
πππππ π ππππππ πππ π πππ
− π ππ ππππππππ πππ
ππ ππ π = π
-5
c.
π(π) =
ππ − π
π+π
π(π) =
ππ − π
ππ π = −π
π+π
(−π)π − π π
π(−π) =
=
−π + π
π
π(π) ππ πππ
ππππππ
ππ π = −π
ππ π = −π
5
π(π) =
y
4
ππ −π
π(π) = π+π ππ π
ππππππππππππ ππ π = −π
π → −π− π → −π
3
2
π
−π. π
−π. ππ
−π. πππ
π(π)
−π. π
−π. ππ
−π. πππ
1
x
-5
-4
-3
-2
-1
1
-1
2
3
4
5
π → −π+ π → −π
-2
π
−π. π
−π. ππ
−π. πππ
-3
π(π)
−π. π
−π. ππ
−π. πππ
-4
-5
ππ − π
πππ πππππ π
ππππππππππππ ππ π = −π
π+π
πππππ π πππππ ππ − π ππ ππππππππ πππ
ππ ππ π = −π
π(π) =
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Name: _________________________________________________ Period: ___________ Date: ________________
Continuity, End Behavior, and Limits Guided Notes
d.
π(π) =
π
πππ
π
ππ π = π
πππ
π
π(π) =
= ∞
π ∗ ππ
π(π) ππ πππ
ππππππ
ππ π = π
π
π(π) = π ππ π
ππππππππππππ ππ π = π
ππ
π → π− π → +∞
ππ π = π
5
π(π) =
y
4
3
2
1
x
-5
-4
-3
-2
-1
1
2
3
4
5
-1
π
−π. π
−π. ππ
−π. πππ
π(π)
ππ. ππ
π, πππ. ππ
πππ, πππ. ππ
π
π. π
π. ππ
π. πππ
π(π)
ππ. ππ
π, πππ. ππ
πππ, πππ. ππ
π → π+ π → +∞
-2
-3
-4
π
πππ ππππππππ π
ππππππππππππ ππ π = π πππππ
πππ
π πππππ ππ + ∞ ππππ π → π
-5
π(π) =
Intermediate Value Theorem
If π(π)is a continuous function and π < π and there is a value π such that π is between π(π) and π(π) , then
there is a numberπ, such that π < π < π and π(π) = π
The Location Principle
If π(π) is a continuous function and π(π) and π(π) have opposite signs, then there exists at least one value π,
such that π < π < π and π(π) = π. That is, there is a zero between π and π.
Sample Problem 2: Determine between which consecutive integers the real zeros of function are located on the given
interval.
a.
π(π) = (π − π)π − π
5
[π, π]
y
4
π
π
π
π
π
π
π
π
π
π
π
−π
−π
−π
π
π
3
2
1
x
-5
-4
-3
-2
-1
1
2
-1
3
4
5
π(π) is positive and π(π) is negative,
π(π) ππππππ ππππ ππ π ≤ π ≤ π
π(π) is negative and π(π) is positive
π(π) ππππππ ππππ ππ π ≤ π ≤ π
-2
-3
-4
π(π) has zeros in intervals:
π ≤ π ≤ π πππ
π ≤ π ≤ π
-5
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Name: _________________________________________________ Period: ___________ Date: ________________
Continuity, End Behavior, and Limits Guided Notes
End Behavior
The end behavior of a function describes what the π -values do as |π| becomes greater and greater.
When π becomes greater and greater, we say that π approaches infinity, and we write π → +∞.
When π becomes more and more negative, we say that π approaches negative infinity, and we write π → −∞.
The same notation can also be used with π or π(π) and with real numbers instead of infinity.
Left - End Behavior (as π becomes more and more negative): π₯π’π¦ π(π)
π→−∞
Right - End Behavior (as π becomes more and more positive): π₯π’π¦ π(π)
π→+∞
The π(π) values may approach negative infinity, positive infinity, or a specific value.
Sample Problem 3: Use the graph of each function to describe its end behavior. Support the conjecture numerically.
a.
π(π) = −πππ + ππ − π
From the graph, it appears that:
π(π) → ∞ as π → −∞ and π(π) → −∞ – as π → ∞
y
8
7
6
The table supports this conjecture.
5
4
3
π
2
1
x
-9 -8 -7 -6 -5 -4 -3 -2 -1-1
1
2
3 4
5
6
7 8
−πππ
π π ∗ ππππ
−πππ
π
πππ
π ∗ πππ
−π
-π ∗ πππ
πππ
-π ∗ ππππ
9
-2
-3
-4
-5
-6
-7
-8
b.
π(π) =
ππ − π
π−π
From the graph, it appears that:
π(π) → −π as π → −∞ and π(π) → −π as π → ∞
y
12
10
The table supports this conjecture.
8
6
4
π
2
-14 -12 -10 -8
-6
-4
-2
x
2
4
6
8
10
12
−πππ
π −π. ππππ −π. ππππ
14
-2
-4
-6
-8
-10
-12
Copyright © PreCalculusCoach.com
−πππ
4
π
πππ
πππ
−π. ππ
−π. πππ
−π. ππππ
Name: _________________________________________________ Period: ___________ Date: ________________
Continuity, End Behavior, and Limits Guided Notes
Increasing, Decreasing, and Constant Functions
A function π is increasing on an interval π° if and only if for every π and π contained in π°, (π) < π(π) , whenever π < π .
A function π is decreasing on an interval π° if and only if for every π and π contained in π°, π(π) > π(π) whenever π < π .
A function π remains constant on an interval π° if and only if for every π and π contained in π°, π(π) = π(π) whenever
π<π .
Points in the domain of a function where the function changes from increasing to decreasing or from decreasing to
increasing are called critical points.
Sample Problem 4: Determine the interval(s) on which the function is increasing and the interval(s) on which the
function is decreasing.
a.
π(π) = ππ − ππ + π
5
From the graph, it appears that:
A function ππ − ππ + π is decreasing for π < π. π
A function ππ − ππ + π is increasing for π > π. π
y
4
3
2
The table supports this conjecture.
1
x
-5
-4
-3
-2
-1
1
2
3
4
5
-1
-2
π
−π
π
π
π
π
π
π
π. π
π
−π. ππ −π. π
π
π
-3
-4
-5
b.
π
π
From the graph, it appears that:
π
A function ππ − π ππ − πππ + π is increasing:
π < −π. ππ πππ
π > π
π
A function ππ − ππ − πππ + π is decreasing:
π
−π. ππ < π < π
π(π) = ππ − ππ − πππ + π
y
12
10
8
6
The table supports this conjecture.
4
2
-14 -12 -10 -8
-6
-4
-2
x
2
4
6
8
10
12
14
π
−π
−π. ππ
−π
π
π
π
π
ππ
ππ. ππ
ππ. π
π
−ππ
−π. π
-2
-4
-6
-8
-10
-12
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