To determine the coordinate point that will have an open circle in the given piecewise function, we need to analyze the two cases provided:
- \( f(x) = 2x + 2 \) for \( x > 4 \)
- \( f(x) = 2x + 1 \) for \( x \leq 2 \)
An open circle on a graph typically indicates a point where the function is not defined. In this case, we need to check the value at the boundary where the two cases meet, which is at \( x = 2 \).
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For \( x = 2 \):
- Since \( x \leq 2 \), we use the second equation: \[ f(2) = 2(2) + 1 = 4 + 1 = 5 \]
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For \( x = 4 \):
- Since \( x > 4 \), we use the first equation. However, we have to check the value directly at \( x = 4 \): \[ f(4) = 2(4) + 2 = 8 + 2 = 10 \]
Since \( f(x) \) is defined when \( x = 4\) with value 10, and we don’t use the equation \( f(x) = 2x + 2 \) for \( x=2\) at that boundary, we need to determine if there is an open circle at \( x=2\).
At \( x=2\), the function value is \( f(2) = 5 \), but the definition only applies for \( x \leq 2\). This indicates that the function does not include the point for the next boundary value before it switches at \( x \) values just over the threshold \( x = 2 \).
Thus, the coordinate point with an open circle will occur at \( (2, 5) \).
So the answer is:
(2, 5)