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The symbol $\lfloor x \rfloor$ denotes the largest integer which is less than or equal to $x$. Thus, $\lfloor 2.1 \rfloor = \lfloor 2 \rfloor = 2$ and $\lfloor -0.9\rfloor = \lfloor-1\rfloor = -1$.

At which points is the following function continuous? $$f(x) = \lfloor \tfrac{1}{x+1} \rfloor$$

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    By "1/x+1", do you mean $\tfrac{1}{x}+1$ or $\tfrac{1}{x+1}$? – 2017-01-06
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    I meant the second one: 1/(x+1). – 2017-01-06
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    Dear camryn. When you ask a question, and you obtain helpful answers, you can upvote as many answers as you'd like to each question you ask. And for each question you ask, you can accept exactly one answer, by clicking on the greyed out $\checkmark$ to the left of the answer you'd like to accept. You get $2$ reputation points whenever you accept one answer to each question you ask. To upvote an answer, and you can upvote as many answers as you receive, click on the "up-arrow" to the left of an answer. (I think the uparrow turns red when you upvote.) – 2017-03-13

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Hint: the function $\lfloor x \rfloor$ is discontinuous at integer values of $x$ and continuous elsewhere, so:

  • $\lfloor \tfrac{1}{x}+1 \rfloor$ is discontinuous if $\tfrac{1}{x}+1$ is an integer and continuous elsewhere;
  • $\lfloor \tfrac{1}{x+1} \rfloor$ is discontinuous if $\tfrac{1}{x+1}$ is an integer and continuous elsewhere.

Note: you probably meant one of these two (see comment), but both are interesting to consider.

Addition after comment

So you meant $\lfloor \tfrac{1}{x+1} \rfloor$ and (see above) you want to know when $\tfrac{1}{x+1}$ is an integer, so when

$$\frac{1}{x+1} =k$$

for some $k \in \mathbb{Z}$. Solve for $x$ to see when this happens.

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    OK so i did that and i got x=-1+ 1/k. But doesn't the answer have to be like -1? – 2017-01-06
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    The function is undefined at $x=-1$, because the denominator is $0$ there. The function is defined at all other values of $x$ and more specifically it is discontinuous where $x = -1+\tfrac{1}{k}$ for $k \in \mathbb{Z}$. – 2017-01-06
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    oh thank u so much. So following that idea if f(x) = [2x] then it would be continuous everywhere right? – 2017-01-06
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    No, it would be continuous everywhere _except_ where $2x$ is an integer, so where $2x=k$ for some $k \in \mathbb{Z}$. – 2017-01-06