To determine which points will have open circles in the graph of the piecewise function
\[ f(x) = \begin{cases} \frac{1}{2}x + 2 & \text{if } x < -3 \ -6 & \text{if } -3 \leq x < 3 \ -x + 2 & \text{if } x \geq 3 \end{cases} \]
we need to analyze the function at the points where the function changes.
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At \(x = -3\):
- For \(x < -3\), the function is given by \(\frac{1}{2}x + 2\). If we substitute \(x = -3\) into the equation: \[ f(-3) = \frac{1}{2}(-3) + 2 = -\frac{3}{2} + 2 = \frac{1}{2} \]
- For \(-3 \leq x < 3\), the function value is \(f(x) = -6\).
- Thus, at \(x = -3\), it transitions from \(f(-3) = \frac{1}{2}\) to \(f(-3) = -6\). This means there will be an open circle at \((-3, \frac{1}{2})\).
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At \(x = 3\):
- For \(x < 3\), \(f(x) = -6\) (constant for \(-3 \leq x < 3\)).
- For \(x \geq 3\), \(f(x) = -x + 2\). If we substitute \(x = 3\): \[ f(3) = -3 + 2 = -1. \]
- The function jumps from \(-6\) to \(-1\) at this point, so there will be an open circle at \((3, -6)\).
From evaluating these points, we can conclude that the following points have open circles:
- \((-3, \frac{1}{2})\)
- \((3, -6)\)
Now, we check the provided options for matching these:
- (−3,12), (3,−1), and (3,−6) – no, as \((-3, 12)\) is not valid and \((3, -1)\) should not be filled in.
- none of the points – no, as some points do have open circles.
- (−3,12), (3,−1), and (3,−6) – no, as \((-3, 12)\) is incorrect.
- (−3,12) and (3,−6) – incorrect for the same reason as above.
Thus, the first statement should have been considered but isn't accurately represented with provided options. Solution implies possible consideration:
- Open circles are at points (-3, 1/2) and (3, -6).
Final answer likely would be “none of the points” in absence of correct options.
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