To determine if the graph accurately represents the piecewise function described, let's analyze the function and the implication of its segments:
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For \(1 \leq x \leq 3\): The dosage is a constant \(0.5\) ounces. This means the graph will have a horizontal line segment from \(x = 1\) to \(x = 3\) with closed circles at both \(x = 1\) and \(x = 3\) since both boundaries are inclusive.
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For \(3 < x \leq 6\): The dosage is \(1.0\) ounces. Therefore, the graph will have another horizontal line segment from just above \(x = 3\) to \(x = 6\) with an open circle at \(x = 3\) (since it is not included) and a closed circle at \(x = 6\).
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For \(x > 6\): The dosage is \(0.5x\). This will be a line that starts at \(x = 6\) (with some initial value \(3\) if you substitute \(6\) into \(0.5x\)) and continues increasing as \(x\) increases. For this portion of the graph, there will be a closed circle at \(x = 6\).
Given this analysis, we can conclude:
- The graph should have closed circles at \(x = 1\) and \(x = 6\).
- There should be an open circle at the point corresponding to \(x = 3\) to correctly represent that \(3\) itself is not included in the second piece of the function.
- The graph is not continuous; it will have a jump between the segments of the graph at \(x = 3\).
Therefore, the correct response would be:
No. There should be an open circle on the first point of two parts of the function.