INCREASING AND DECREASING FUNCTIONS
SECTION 1.4:
CONTINUITY AND ONE-SIDED LIMITS
Goals: The Student Will Be Able To:
❖ Define and apply the concepts of Continuity at a Point, Continuity on an Open Interval and Continuity on a Closed Interval
❖ Recognize, understand and distinguish between Nonremovable and Removable discontinuities
❖ Discuss the continuity of various functions, and corresponding graphical implications
❖ Differentiate between infinite and jump Nonremovable discontinuities
❖ Evaluate limits from the left and right (one-sided limits)
❖ Understand, explain and apply the Intermediate Value Theorem
DEFINITION OF CONTINUITY
Continuity at a Point:
A function is continuous at “c” if each of t he these three (3) conditions are met:
1. [pic]is defined
2. [pic]exists
3. [pic]
Continuity on an Open Interval:
A function is continuous on an open interval (a, b), if it is continuous at every point within the interval.
Continuity Everywhere:
A function is continuous everywhere, if (a, b) is[pic].
EXAMPLES OF DISCONTINUITY
2 TYPES OF DISCONTINUITY-
REMOVABLE AND NONREMOVABLE
REMOVABLE: If the discontinuity can be taken away by defining or redefining a value for the function. “Fill in the hole, or erase the point and move it to the hole.”
NONREMOVABLE: If the discontinuity cannot be taken away. Defining or redifinig a point will not reconnect the function. “JUMP” and “INFINITE” discontinuities are two basic types.
EXAMPLE 1:
Discuss the continuity of each function:
a. [pic]
Since at x = 0, the function is not defined, there is a discontinuity at x = 0. Therefore f is continuous for [pic]. Considering the familiar graph we can see it is a NONREMOVABLE – INFINTE type.
b. [pic]
Again, we can see that x =1 will produce a 0 in the denominator. All other x values are in the domain, so f is continuous for [pic]. However, due to analytic cancellation, [pic]is a function identical to f except for the “hole” at (1, 2). Since the discontinuity is just a “hole”, this is the REMOVABLE type.
c. [pic]
This is a special type of function which joins more than 1 equation. 2 separate pieces of graphs are used. f is called a “PECEWISE” function. Since thepieces join, and do not “JUMP”, the function is CONTINUOUS for all x … [pic].
d. [pic]
The domain for this trigonometric function is all real numbers. There are no discontinuities at all. So the interval of continuity is [pic].
ONE-SIDED LIMITS
We have already suggested that it is necessary to examine a neighborhood both to the left and to the right of “c”, when evaluating a limit. Now we will give it a formal name and formal notation.
“Limit from the right” ( [pic]
Examine the function for values greater than c … or the “positive side of” c.
“Limit from the left” ( [pic]
Examine the function for values less than c … or the “negative side of” c.
EXAMPLE 2:
Find the limit of [pic]as x approaches –2 from the right.
With either numerical analysis, or simple visual inspection we can see the function is approaching 0, as x approaches –2 from the right.
.EXAMPLE 3:
Find the limit of the “Greatest Integer Function” [pic] as x approaches 0 from the left and from the right.
Since [pic], or no matter how close x gets to 0 from the right, f is still 0 …[pic]
Yet as x gets close to 0 from the left,
The value of the function remains
a steady –1. Ex: [pic]
So then [pic]
We can conclude that the “Greatest Integer Function” is not continuous at ANY integer value. These are all JUMP discontinuities
THE EXISTENCE OF A LIMIT
Let f be a function and let c and L be real numbers. The limit as x approaches c exists if and only if:
[pic] and [pic]
CONTINUITY ON A CLOSED INTERVAL
A function f is continuous on an closed interval [a, b], if it is continuous on the open interval (a, b) and:
[pic] and [pic]
Basically, the endpoints need to fill filled-in pieces of the graph.
EXAMPLE 4: Discuss the continuity of [pic]
For this radical function, it is clearly continuous on the open interval (-1, 1).
And since
[pic]… Continuous from the right at –1
[pic] … Continuous from the left at 1
We can conclude that f is continuous on the closed interval
[-1, 1], as this can be visually verified on the graph.
PROPERTIES OF CONTINUITY
If b is a real number and f and g are continuous at x = c, then the following functions are also continuous at x = c.
1. Scalar Multiple: bf 2. Sum and Difference: f ( g
3. Product: f ( g 4. Product: f / g, g ( 0
Two continuous functions can be multiplies by a constant, added, subtracted, multiplied or divided by each other, and still be continuous!
EXAMPLE 7: Describe the intervals on which each function is continuous.
a. [pic]… The tangent function is undefined at [pic], and at repeated intervals [pic]where n is an integer. It is continuous within these discontinuous values. So it is continuous on open intervals:
[pic][pic][pic]
THE INTERMEDIATE VALUE THEOREM
If f is continuous on the closed interval [a, b], and k is any number between f(a) and f(b), then: the exists (() at least one number c in [a, b] such that f(c) = k.
For any theorem to apply, the hypothesis must be true. So to apply the IVT, we must verify/state the truth of the 2 parts of the hypothesis.
1. The function must be continuous on [a, b]
2. k is selected between the two y values … f(a) and f(b)
( 3. Conclusion: There is at least one c value within (a, b) such that f(c) = k. We show 3 c’s in our example above.
EXAMPLE 8:
Use the IVT to show that the function [pic]has a zero in the interval [0, 1].
Solution: To apply the IVT, recall, we must establish the 2-part hypothesis is valid.
1. Since f is a polynomial, it is permissible to assume that it is continuous for all x, and therefore on our interval [0, 1]
2. f(0) = -1 and f(1) = 2, an the k value suggested in the problem is Zero. It is true that 0 is between the two y values –1 and 2.
So, the IVT will apply, and there must be at least 1 c value between 0 & 1 which will yield f(c) = 0
A graphing calculator can be used to find that “c” … the IVT just guarantees its existence. Also, the graph to the right illustrates the value.
-----------------------
a
b
c
(
)
Rule 2:
f(c) is defined, but left and right limits do not match …so limit does not exist
a
b
c
(
)
Rule 1:
f(c) is not defined
a
b
c
(
)
Rule 3:
left and right limits do match …f(c) is defined, but f(c) ( limit
a
b
c
(
)
a
b
c
(
)
a
b
c
(
)
JUMP
c
INFINITE
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
a
b
[
]
k
f(a)
a
b
c1
[
]
f(b)
k
c2
c32
f(a)
f(b)
Theorem Applies
Theorem Doesn’t Apply
[pic]
c
k
................
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