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 look at the behavior of the function at \( x = 1 \).
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For \( x < 1 \): The function is defined as \( f(x) = -x - 1 \). At \( x = 1 \), we can calculate \( f(1) \) using this piece: \[ f(1) = -1 - 1 = -2. \] However, this piece is not valid at \( x = 1 \), because the condition requires \( x \) to be strictly less than 1.
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For \( x \geq 1 \): The function is defined as \( f(x) = -x + 2 \). We can calculate this at \( x = 1 \) as well: \[ f(1) = -1 + 2 = 1. \] Here, this piece is valid at \( x = 1 \).
Since the function value for the piece where \( x < 1 \) does not apply at \( x = 1 \), we will place an open circle at the point \( (1, -2) \) indicating that this value is not included in the graph.
To summarize, the point that should have an open circle is \( (1, -2) \).
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