To determine where the open circle will be located for the piecewise function given by:
\[ f(x) = \begin{cases} 2x + 2 & \text{if } x > 2 \ 2x + 1 & \text{if } x \leq 2 \end{cases} \]
we need to evaluate the function at the point where the definition of the function changes, which is at \(x = 2\).
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Evaluate \( f(2) \) using the second piece (since it is defined for \(x \leq 2\)): \[ f(2) = 2(2) + 1 = 4 + 1 = 5 \]
So, when \(x = 2\), \(f(2) = 5\).
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For \(x > 2\), we switch to the first piece of the function. We can check the limit as \(x\) approaches 2 from the right: \[ \lim_{x \to 2^+} f(x) = 2(2) + 2 = 4 + 2 = 6 \]
Thus:
- At \(x = 2\), the function \(f(x)\) is defined and equals 5, indicating a filled circle at the point \((2, 5)\).
- Approaching from the right, the function's value approaches 6, which means there will be an open circle at \((2, 6)\) because the value of the function for \(x > 2\) is not defined at \(x = 2\) (it takes the value of 5 from the left, indicated by the closed circle, while approaching the value 6 from the right).
Therefore, the coordinate point that will have an open circle is:
\[ \boxed{(2, 6)} \]
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