Question

which statement best compares the graphs of $f(x)=\lfloor x \
floor$ and $f(x)=\lceil x \
ceil$?
\bigcirc\\ the two graphs are exactly the same.
\bigcirc\\ the open and closed circles are reversed.
\bigcirc\\ the graph of $f(x)=\lceil x \
ceil$ is shifted up compared to the graph of $f(x)=\lfloor x \
floor$, and the open and closed circles are reversed.
\bigcirc\\ the graph of $f(x)=\lceil x \
ceil$ is shifted up compared to the graph of $f(x)=\lfloor x \
floor$, and the open and closed circles are the same.

Explanation:

1. Recall the definitions: The floor function \( \lfloor x

floor \) gives the greatest integer less than or equal to \( x \), and the ceiling function \( \lceil x
ceil \) gives the least integer greater than or equal to \( x \).

1. For a value like \( x = 2.3 \), \( \lfloor 2.3

floor=2 \) and \( \lceil 2.3
ceil = 3 \), so the ceiling function's graph is shifted up by 1 unit relative to the floor function in non - integer intervals.

1. Regarding the open and closed circles: For \( \lfloor x

floor \), at an integer \( n \), the closed circle is at \( (n, n) \) (since \( \lfloor n
floor=n \)) and the open circle is at \( (n, n - 1) \) (as \( x\) approaches \( n\) from the right, \( \lfloor x
floor\) approaches \( n- 1\) for \( x\in(n - 1,n)\)). For \( \lceil x
ceil \), at an integer \( n \), the closed circle is at \( (n, n) \) (since \( \lceil n
ceil=n \)) and the open circle is at \( (n, n + 1) \) (as \( x\) approaches \( n\) from the left, \( \lceil x
ceil\) approaches \( n + 1\) for \( x\in(n,n + 1)\)). So the open and closed circles are reversed between the two functions, and the ceiling function's graph is shifted up compared to the floor function.

Answer:

C. The graph of \( f(x)=\lceil x
ceil \) is shifted up compared to the graph of \( f(x)=\lfloor x
floor \), and the open and closed circles are reversed.