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Find the discontinuity at $f(2)$ of the function $f(x)=\dfrac{x^2-3x+2}{{x^2}+x-6}$.

I am confused. I do not understand that is there discontinuity at $2$ but it has discontinuity at $x=-2$. can you explain it please? For my point of view, there is no discontinuity at $2$ because after factorization I get $f(2)= $1/5 and before factorization I get $f(2)=0$ as well...

robax
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    A rational function is continuous on its domain. What is the domain of $f$? – user251257 Jan 07 '17 at 15:46
  • How did you got $f(2)=0$ ? – curious_mind Jan 07 '17 at 15:48
  • just plug x=2 into the main function – robax Jan 07 '17 at 15:49
  • Sorry for previous comment, it was before the edit. – curious_mind Jan 07 '17 at 15:49
  • And actually, $f(2)$ is not defined. I mean, $2 \notin D_f$. – curious_mind Jan 07 '17 at 15:51
  • Most likely, what the question wants is that you calculate what value to "give" your function to bridge the "gap" at $x=2$. – sTertooy Jan 07 '17 at 15:54
  • you are right but if you draw the function at x=2, you get y=0; but x=3 it is continuous. how you draw the function. because between f(2) and F(-1) and f(3) function changes its sign from minus to plus – robax Jan 07 '17 at 15:55
  • you are right @SteamyRoot but my question is that is there discontinuity at x=2 if yes , why? if you draw the function how you show the discontinuity? – robax Jan 07 '17 at 16:00
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    @robax, if you mean $y=f(x)$, you can't get $y=0$ by putting $x=2$ in main function because $2 \notin D_f$. Another thing, if you want to "fill the gap", you should assign $f(2)=\dfrac{1}{5}$. If you will do this, the function will become continuous. – curious_mind Jan 07 '17 at 16:00
  • sorry, I do not mean this. I do not like to fill up discontinuity. I like to see how you show discontinuity on the graph. – robax Jan 07 '17 at 16:02
  • @robax There is no discontinuity because your function does not exist there. In every $x$ where $f(x)$ is defined, $f$ will be continuous. – sTertooy Jan 07 '17 at 16:56
  • This function has a removable discontinuity at $x=2.$ By defining $f(2) = \frac 1 5$ you can make $f$ continuous at $x=2.$ But the discontinuity at $x=3$ is not removable and in fact $f$ has an infinite discontinuity at $x=-3.$ Observe that $f(x) \to +\infty$ as $x \to -3^{-}$ and $f(x) \to -\infty$ as $x \to -3^+.$ – math maniac. Jul 03 '20 at 04:37

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Remember that a rational function is continuous across its domain. That is, it is continuous wherever it is defined. We can see that the function $f(x)$ is undefined whenever the denominator is $0$ as division by zero is undefined. You claim in your question that $f(x)$ is discontinuous at $x=-2$; I assure you that it is not. Because $f(-2)$ is defined, $x=-2$ is in the domain of this function and is therefore continuous there. $$f(-2)=\frac{(-2)^2-3(-2)+2}{(-2)^2+(-2)-6}=\frac{4+6+2}{4-2-6}=\frac{12}{-4}=-3$$ We can simplify $f(x)$ only if $x\ne2$. This is because the denominator will become $0$ if $x=2$. $$f(x)=\frac{x^2-3x+2}{x^2+x-6}=\frac{(x-2)(x-1)}{(x-2)(x+3)}$$ Notice what happens if we let $x=2$. $$f(2)=\frac{((2)-2)((2)-1)}{((2)-2)((2)+3)}=\frac{0\cdot1}{0\cdot5}=\frac{0}{0}$$ This is undefined (it is also an indeterminate form). Notice how you obtained $1/5$ here. By simply removing $(x-2)$ from both the numerator and denominator, you essentially removed $0/0$ when $x=2$. You cannot do this. You will obtain confusing and erroneous results when you mistakenly divide by zero. This is the main point of your confusion here. Now, let us determine what type of discontinuity exists at $x=2$ as the problem asks. We know that $f(2)$ is undefined, but we can evaluate the limit as $x\to2$ to see what is happening there. We can remove $(x-2)$ when evaluating the limit because $x$ is only approaching $2$; it will not equal it, so we won't have a problem. That is what a limit implies. We can use direct substitution. $$\lim\limits_{x\to2}(f(x))=\frac{(2)-1}{(2)+3}=\frac{1}{5}$$ Because $f(2)$ is undefined and the limit as $x\to2$ exists, we can then say that there exists a removable discontinuity at $x=2$. This is sometimes called a "hole" because of a missing point in the graph of the function where $(2,f(2))$ should be. If $f(x)$ were to be continuous at $x=2$, it would need to be defined there and equal to this limit. Thus, $f(2)$ would need to be equal to $1/5$. In all, the function $f(x)$ is continuous on $(-\infty,-3)\cup(-3,2)\cup(2,\infty)$ as that is the domain of the function. Alternatively, you could say there is a removable discontinuity at $x=2$ because we could "remove" the expression $(x-2)$ from the function as it would have lead to a discontinuity at a certain value (if $x=2$). This intuition, of course, is informal and simply pedagogical. The same cannot be said for $x=-3$, however. The function is also discontinuous there, but it is a different type of discontinuity—an infinite discontinuity.

ThunderGod763
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This function has a removable discontinuity at $x=2.$ By defining $f(2) = \frac 1 5$ you can make $f$ continuous at $x=2.$ But the discontinuity at $x=-3$ is not removable and in fact $f$ has an infinite discontinuity at $x=-3.$ Observe that $f(x) \to +\infty$ as $x \to -3^{-}$ and $f(x) \to -\infty$ as $x \to -3^+.$ $f$ is continuous at every other real number.

math maniac.
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