Question

which is the graph of the step function f(x)?
$f(x)=\

$$\begin{cases}-1, & x < -1 \\\\ 0, & -1 \\leq x \\leq 1 \\\\ 1, & x > 1\\end{cases}$$

$

Explanation:

Step1: Analyze \( x < -1 \)

For \( x < -1 \), \( f(x) = -1 \). So the graph should be a horizontal line at \( y = -1 \) for all \( x \) values less than -1. Check the graphs: the first graph has a horizontal segment at \( y = -1 \) (left of \( x = -1 \)), which matches. The second graph doesn't show this segment.

Step2: Analyze \( -1 \leq x \leq 1 \)

For \( -1 \leq x \leq 1 \), \( f(x) = 0 \). So a horizontal line at \( y = 0 \) between \( x = -1 \) (inclusive) and \( x = 1 \) (inclusive). The first graph has a segment at \( y = 0 \) here (closed circles at -1 and 1? Wait, the function for \( -1 \leq x \leq 1 \) is 0, so the segment should cover from \( x = -1 \) (inclusive) to \( x = 1 \) (inclusive). The first graph's middle segment (around \( x \) from -1 to 1) is at \( y = 0 \)? Wait, looking at the first graph: left segment ( \( x < -1 \)) is \( y = -1 \), middle segment ( \( -1 \leq x \leq 1 \))? Wait, maybe the first graph's middle part: wait, the first graph has a segment at \( y = -1 \) (left), then at \( y = 0 \) (middle, from \( x = -1 \) to \( x = 1 \)), then at \( y = 1 \) (right, \( x > 1 \))? Wait, no, the function is \( f(x) = 0 \) for \( -1 \leq x \leq 1 \), so the middle segment should be at \( y = 0 \). The first graph's middle segment (the one between \( x = -1 \) and \( x = 1 \)): looking at the grid, if \( x < -1 \), \( y = -1 \); \( -1 \leq x \leq 1 \), \( y = 0 \); \( x > 1 \), \( y = 1 \). The first graph shows this: left part ( \( x < -1 \)) at \( y = -1 \), middle ( \( -1 \leq x \leq 1 \)) at \( y = 0 \), right ( \( x > 1 \)) at \( y = 1 \). The second graph doesn't have the \( y = -1 \) segment, so it's incorrect.

Step3: Analyze \( x > 1 \)

For \( x > 1 \), \( f(x) = 1 \). So a horizontal line at \( y = 1 \) for \( x > 1 \). The first graph has a segment at \( y = 1 \) (right of \( x = 1 \), open circle? Wait, \( x > 1 \), so open circle at \( x = 1 \), and the segment goes to the right. The first graph's right segment is at \( y = 1 \), which matches. The second graph's right segment is at \( y = 2 \) or something? No, the second graph's right segment is at \( y = 2 \)? Wait, no, the second graph's right part is at \( y = 2 \)? Wait, no, the first graph's right segment is at \( y = 1 \), which matches \( f(x) = 1 \) for \( x > 1 \).

Answer:

The graph that matches is the first one (the upper graph among the two shown, with segments at \( y = -1 \) (left), \( y = 0 \) (middle), and \( y = 1 \) (right)).