Question

representing a graph with a step function
which function and domain could represent the given graph?
$f(x) = 1 - \lfloor x \
floor$; $-3 < x < 3$
$f(x) = 1 + \lfloor x \
floor$; $-3 \leq x \leq 3$
$f(x) = -\lfloor x \
floor$; $-3 \leq x \leq 3$
$f(x) = \lfloor x \
floor$; $-3 < x < 3$

Explanation:

Step1: Analyze the graph's key points

The graph has points like at \( x = 0 \), \( y = 0 \); at \( x = 1 \), \( y=-1 \); at \( x = -1 \), \( y = 1 \); at \( x = 2 \), \( y=-2 \); at \( x=-2 \), \( y = 2 \); at \( x = 3 \), \( y=-3 \); at \( x=-3 \), \( y = 3 \). Let's check each function:

Step2: Check \( f(x)=1 - \lfloor x

floor \) (Wait, no, the options have \( \lfloor x
floor \)? Wait, the options have \( |x| \)? Wait, maybe it's the floor function? Wait, no, the options are with absolute value? Wait, no, maybe it's the greatest integer function? Wait, the options are:

1. \( f(x)=1 - \lfloor x

floor \); \( -3 < x < 3 \)

1. \( f(x)=1 + \lfloor x

floor \); \( -3 \leq x \leq 3 \)

1. \( f(x)=-\lfloor x

floor \); \( -3 \leq x \leq 3 \)

1. \( f(x)=\lfloor x

floor \); \( -3 < x < 3 \)

Wait, let's re-express. Let's take \( x = 0 \):

For option 3: \( f(0)=-\lfloor 0
floor = 0 \), which matches the graph (at \( x=0 \), \( y=0 \)).

At \( x = 1 \): \( f(1)=-\lfloor 1
floor = -1 \), matches (at \( x=1 \), \( y=-1 \)).

At \( x=-1 \): \( f(-1)=-\lfloor -1
floor = -(-1)=1 \), matches (at \( x=-1 \), \( y=1 \)).

At \( x=2 \): \( f(2)=-\lfloor 2
floor = -2 \), matches (at \( x=2 \), \( y=-2 \)).

At \( x=-2 \): \( f(-2)=-\lfloor -2
floor = -(-2)=2 \), matches (at \( x=-2 \), \( y=2 \)).

At \( x=3 \): \( f(3)=-\lfloor 3
floor = -3 \), and the domain is \( -3 \leq x \leq 3 \), so \( x=3 \) is included, and the graph has a point at \( x=3 \), \( y=-3 \), which matches.

At \( x=-3 \): \( f(-3)=-\lfloor -3
floor = -(-3)=3 \), matches the graph's point at \( x=-3 \), \( y=3 \).

Now check other options:

Option 1: \( f(0)=1 - \lfloor 0
floor = 1 \), doesn't match \( y=0 \) at \( x=0 \).

Option 2: \( f(0)=1 + \lfloor 0
floor = 1 \), doesn't match.

Option 4: \( f(0)=\lfloor 0
floor = 0 \), but \( f(1)=\lfloor 1
floor = 1 \), which doesn't match \( y=-1 \) at \( x=1 \).

So the correct function is \( f(x)=-\lfloor x
floor \) with domain \( -3 \leq x \leq 3 \). Wait, but the options given in the problem (maybe a typo, but based on the graph, the third option is \( f(x)=-\lfloor x
floor ; -3 \leq x \leq 3 \))? Wait, the original problem's options:

Wait the user's problem has options:

• \( f(x)=1 - \lfloor x

floor ; -3 < x < 3 \)

• \( f(x)=1 + \lfloor x

floor ; -3 \leq x \leq 3 \)

• \( f(x)=-\lfloor x

floor ; -3 \leq x \leq 3 \)

• \( f(x)=\lfloor x

floor ; -3 < x < 3 \)

So the third option is \( f(x)=-\lfloor x
floor ; -3 \leq x \leq 3 \), which matches the graph.

Answer:

\( f(x)=-\lfloor x
floor ; -3 \leq x \leq 3 \) (the third option: \( f(x)=-\lfloor x
floor ; -3 \leq x \leq 3 \))