Continuity & Differentiability
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Part B: Differentiability Exercises for This Topic Index of On-Line Topics Everything for Calculus Everything for Finite Math Everything for Finite Math & Calculus Utility: Function Evaluator & Grapher Español |
Part A: Continuity
Note To understand this topic, you will need to be familiar with limits, as discussed in the chapter on derivatives in Calculus Applied to the Real World. If you like, you can review the topic summary material on limits or, for a more detailed study, the on-line tutorial on limits.
To begin, we recall the definition (see Section 6 in the derivatives chapter in Calculus Applied to the Real World.) of what it means for a function to be continuous at a point or on a subset of its domain.
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Continuous at a Point
The function f is continuous at the point a in its domain if:
Note
Continuous on a Subset of the Domain
Examples
2. The function f(x) = 1/x, also of closed form, is continuous at every point of its domain. (Note that 0 is not a point of the domain of f, so we don't discuss what it might mean to be continous or discontinuous there.) 3. The function
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Example 1 Recognizing Points of Discontinuity in a Graph
Let f have the graph shown below.
At which points of the domain of f is the function f discontinuous?
Solution
According to the definition, f can fail to be continuous at a point a of its domain if either:
| 1. | lim x→a | f(x) does not exist, or |
| 2. | lim x→a | f(x) exists, but ≠ f(a) |
Looking at the figure, we see that the possible points where things might go wrong are at x = -1, 0, 1, and 2. Let us look at these points one-at-a-time.
x = -1: Notice that x = -1 is a point of the domain of f, and lim x→-1f(x) does not exist, because the left- and right-habd limits disagree. Therefore, f has a discontinuity at x = -1.
x = 0: From the graph, we see that f(0) is not defined. Therefore, 0 is not in the domain of f, so we cannot say that f has a discontinuity at x = 0. (Some authors would say that f is discontinuous at x = 0, but we don't consider such points...)
x = 1: At the point where x = 1, the limit does exist, however,
| lim x→1 | f(x) = 1, but |
| f(1) | = 2 |
x = 2: Although there is a cusp in the graph at x = 2, you will see that the limit at x = 2 exists, and agrees with f(2). Therefore, f is continuous at x = 2.
Thus, the only points of discontinuity in the domain of f occur at x = -1 and x =1.
Here is a more interactive example.
Example 2 Recognizing Points of Discontinuity in a Graph
Let f have the graph shown below.
The next examples discuss functions specified algebraically rather than geometrically.
Example 3 Identifying Points of Discontinuity in a Non-Closed Form Function
Let f be specified as
| f(x) = |  | x2+4 x + 1 x2 + 1 | if x ≤ 0 if 0 < x ≤ 1 if x > 1 |
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Looking near x = 0, we get
| lim x→0- | f(x) = 4 | Substitute x = 0 in the first (closed-form) formula |
| lim x→0+ | f(x) = 1 | Substitute x = 0 in the second (closed-form) formula |
Since these limits disagree, limx→0f(x) does not exist, and so the function is discontinuous at x = 0.
If you now compute the left- and right limits at x = 1, on the other hand, you will find they agree and equal 2, which is also the value of f(1). Thus, f is continuous at x = 1, and so the only point of discontinuity is x = 0.
Example 4 Fixing up a Function
Let f be given by
| f(x) = | | x2 - x + 1 kx2 + 1 | if x ≤ 1 if x > 1 |
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Solution
Since the only place where something can go wrong is at x = 1, we look at the left- and right limits, as well as the value of the function.
| lim x→1- | f(x) = 1 |
| lim x→1+ | f(x) = k+1 |
| f(1) = 1 | |
You can now either go on and try those exercises that deal with continuity in the exercise set for this topic, or first finish the on-line text by going to Part B: Differentiability.
