I noticed something on wolframalpha : here
Why is this function: $$ \frac{x+3}{(x+3)(x-8)} $$ not defined at $x=-3$?
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If it is a "rational function" then it is defined at $, x = -3.,$ Perhaps that's what you are thinking of? – Bill Dubuque Dec 25 '13 at 17:56
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1For some purposes one would take it to be defined at $x=-3$, by finding $1/(x-8)$ when $x=-3$. It's called a "removable discontinuity". – Michael Hardy Dec 25 '13 at 18:03
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So what is the correct answer? Is it discontinuous, is it not? Doesn't discontinuous mean infinite value at that point? (except at points where the function "jumps" finite values) – BinaryBurst Dec 25 '13 at 19:32
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The (x+3) at the bottom. – Shahar Dec 25 '13 at 19:37
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The answer depends on what type of "function" the expression denotes, which you have not precisely specified. – Bill Dubuque Dec 25 '13 at 19:57
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Related : http://math.stackexchange.com/questions/426437/finding-the-range-of-fx-1-x-1x-2 http://math.stackexchange.com/questions/472169/find-extreme-values-of-frac2xx4 http://math.stackexchange.com/questions/481381/prove-that-forall-x-in-bbb-r-0-lt-frac1-x26x10-le-1?lq=1 http://math.stackexchange.com/questions/533021/what-is-the-supremum-and-infimum-of-n-1n2-where-n-is-an-element-of-mat?lq=1 http://math.stackexchange.com/questions/443322/what-is-the-maximum-value-of-frac2xx-1-fracxx-1-if-x-in-ma http://math.stackexchange.com/questions/577148/range-of-this-function-fracx2-x-1-x-4 – lab bhattacharjee Dec 26 '13 at 04:36
4 Answers
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The function, as given, is undefined at both $x = -3$ and $x = 8$, because the denominator of your function is $0$ at $x = -3$, just as the denominator is $0$ at $x = 8$, and I'm sure you recall that division by zero is undefined.
$x = -3$ is called a removable discontinuity, (better referred to as a "removable singularity") so you can cancel the common factor $(x+3)$ under the provision that $x\neq -3$, and then graph the function just as you might graph $f(x) = \dfrac 1{x-8},$ but for the given function, as posted, there needs to be a "hole" in the graph where $x = -3$.
amWhy
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but if i take the limit at x=3 of the given function there is no discontinuity and even if i plot the given function there still isn't any discontinuity ... I am extremely confused, I'm starting to doubt everything I know about math... ;( – BinaryBurst Dec 25 '13 at 19:27
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The fact that the limit as $x\to -3$ exists (which is not the same as the function's value at $x = -3,$ since the function is undefined at $x = -3$) (and that the limit) is finite means precisely that $x = -3$ is what we call a removable discontinuity. Compare to $x=8$, where the function is undefined, but where $x = 8$ is NOT a removable discontinuity. When you plot the function, the behavior of the function displays just as it would if we canceled the factor $(x+3)$ and graphed it, except there needs to be a "hole" at $x = 3$. – amWhy Dec 25 '13 at 19:36
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That "hole" simply is not indicated on the graph produced by Wolfram Alpha (one single point that's missing from a function's domain, when it's a removable discontinuity won't typically show up in WA), but technically, the graph needs to be precisely the graph of $\frac 1{x-8}$, except that at $(-3, -1/11)$, we need to draw a tiny open circle to indicate the "hole" at that point. – amWhy Dec 25 '13 at 19:44
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NB: You're thinking of discontinuity as meaning essential discontinuity, which is what is the case at $x = 8$. A removable discontinuity tells us that the behavior of the function as $x \to -3$ in this case is just as it would be if $x = -3$ were in the domain, even though it's not in the domain. – amWhy Dec 25 '13 at 19:49
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This might clarify matters: Removable Discontinuity aka Removable Singularity Read through the entry long enough to read the excerpt about "abuse of terminology" – amWhy Dec 25 '13 at 19:52
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At x = -3, the denominator would become 0 and so will the numerator, thus the entire fraction would become $\frac00$ which is undefined. Thus the function is undefined at x = -3.
user2369284
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The remark "that the entire fraction would become $\infty$" is imprecise, and arguably incorrect, since the apparent singularity is removable; it has finite limit as $,x\to -3.$ – Bill Dubuque Dec 25 '13 at 17:51
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Division is a function of two variables. The domain of $(x,y)\mapsto x/y$ is $$\{(x,y)\mid y \not = 0\}.$$ What do you know about domains and compositions of functions?
ncmathsadist
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You have to be careful when plotting these types of graphs with graphing utilities. Many don't have the capabilities (or insight) to show you the discontinuity. The graph of the function should look like the following image.
Have a look at the different types of discontinuities at http://www.math.brown.edu/UTRA/discontinuities.html. The discontinuity in $f(x)=\dfrac{x+3}{(x+3)(x-8)}$ is the second type mentioned in that file (Point Discontinuity).
Tyler Clark
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