How To Determine Whether A Function Is Continuous Or Discontinuous
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Explore Book Pre-Calculus All-in-One For Dummies
Explore BookBuy NowBuy on AmazonBuy on WileySubscribe on Perlego
Explore Book Pre-Calculus All-in-One For Dummies
Explore BookBuy NowBuy on AmazonBuy on WileySubscribe on PerlegoA graph for a function that's smooth without any holes, jumps, or asymptotes is called continuous. Your pre-calculus teacher will tell you that three things have to be true for a function to be continuous at some value c in its domain: -
f(c) must be defined. The function must exist at an x value (c), which means you can't have a hole in the function (such as a 0 in the denominator).
-
The limit of the function as x approaches the value c must exist. The left and right limits must be the same; in other words, the function can't jump or have an asymptote. The mathematical way to say this is that
must exist.
-
The function's value at c and the limit as x approaches c must be the same.

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is continuous at x = 4 because of the following facts:
-
f(4) exists. You can substitute 4 into this function to get an answer: 8.
If you look at the function algebraically, it factors to this:
Nothing cancels, but you can still plug in 4 to get
which is 8.
Both sides of the equation are 8, so f(x) is continuous at x = 4.
Functions that aren't continuous at an x value either have a removable discontinuity (a hole in the graph of the function) or a nonremovable discontinuity (such as a jump or an asymptote in the graph):
-
If the function factors and the bottom term cancels, the discontinuity at the x-value for which the denominator was zero is removable, so the graph has a hole in it.
For example, this function factors as shown:
After canceling, it leaves you with x – 7. Therefore x + 3 = 0 (or x = –3) is a removable discontinuity — the graph has a hole, like you see in Figure a.
The graph of a removable discontinuity leaves you feeling empty, whereas a graph of a nonremovable discontinuity leaves you feeling jumpy.
-
If a term doesn't cancel, the discontinuity at this x value corresponding to this term for which the denominator is zero is nonremovable, and the graph has a vertical asymptote.
The following function factors as shown:
Because the x + 1 cancels, you have a removable discontinuity at x = –1 (you'd see a hole in the graph there, not an asymptote). But the x – 6 didn't cancel in the denominator, so you have a nonremovable discontinuity at x = 6. This discontinuity creates a vertical asymptote in the graph at x = 6. Figure b shows the graph of g(x).
About This Article
This article is from the book:
Pre-Calculus For DummiesAbout the book author:
Mary Jane Sterling (Peoria, Illinois) is the author of Algebra I For Dummies, Algebra Workbook For Dummies, Algebra II For Dummies, Algebra II Workbook For Dummies, and many other For Dummies books. She taught at Bradley University in Peoria, Illinois for more than 30 years, teaching algebra, business calculus, geometry, and finite mathematics.
This article can be found in the category:
Pre-Calculus
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