Question

Which point should have an open circle in the graph of the piecewise function f(x)={−x−1 if x<1−x+2 if x≥1?(1 point)
Responses

(−1,1)
left parenthesis negative 1 comma 1 right parenthesis

(1,−2)
left parenthesis 1 comma negative 2 right parenthesis

(−1,−2)
left parenthesis negative 1 comma negative 2 right parenthesis

(1,1)
left parenthesis 1 comma 1 right parenthesis
Remaining Attempts : 3
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Answer 1

To determine which point should have an open circle in the graph of the piecewise function \( f(x) = \begin{cases} -x - 1 & \text{if } x < 1 \\ -x + 2 & \text{if } x \geq 1 \end{cases} \), we need to evaluate the function at the point where the definition changes, which is at \( x = 1 \).

1. For \( x < 1 \), the function is \( f(x) = -x - 1 \).
- If we plug in \( x = 1 \) (even though it's not less than 1, we can find what value it approaches):
\[ f(1) = -1 - 1 = -2 \]

2. For \( x \geq 1 \), the function is \( f(x) = -x + 2 \).
- If we plug in \( x = 1 \):
\[ f(1) = -1 + 2 = 1 \]

Since the value of the function changes at \( x = 1 \) — it approaches \(-2\) from the left but is defined as \(1\) for \(x = 1\) from the right — we need to indicate that the point \((1, -2)\) is not included in the graph.

Therefore, the point that should have an open circle in the graph is:
**(1, -2)**.