In the given piecewise function:
\[ f(x) = \begin{cases} -x - 1 & \text{if } x < 1 \ -x + 2 & \text{if } x \geq 1 \end{cases} \]
We need to analyze the points given, particularly how they relate to the transition at \( x = 1 \).
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For \( x < 1 \), the function is defined as \( f(x) = -x - 1 \).
- At \( x = -1 \): \[ f(-1) = -(-1) - 1 = 0 - 1 = -1 \] This point is \((-1, -1)\), but it does not give a direct implication for an open circle.
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For \( x \geq 1 \), the function is \( f(x) = -x + 2 \).
- At \( x = 1 \): \[ f(1) = -1 + 2 = 1 \] This means we have point \( (1, 1) \) for \( x = 1 \).
Now let's check the points you've listed:
- \( (-1, 1) \) is not corresponding to any of the calculated values.
- \( (1, -2) \): This point does not match since \( f(1) = 1 \).
- \( (-1, -2) \): This point is unrelated since it's not a calculated value of the function.
- \( (1, 1) \): This point matches with the calculation \( f(1) = 1 \).
To summarize: at \( x = 1 \), \( f(x) = 1\) is included in the second piece of the piecewise function \( (-x + 2) \) since it specifies \( x \geq 1\).
However, at \( x = 1 \), we have a transition from the first function segment where \( f(1) \) would be calculated as \(-1 - 1 = -2\), but since \( f(1) \) is defined in the second segment, we denote this as having an open circle at \( (1, -2) \).
Therefore, the point that should have an open circle is:
(1, -2)