To find the coordinate point with an open circle in the given piecewise function, let's analyze the function:
( f(x) = \begin{cases} 2x + 2 & \text{if } x > 4 \ \frac{1}{2}x + 1 & \text{if } x \leq 2 \end{cases} )
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Condition for \( x > 4 \): For \( x \) values greater than 4, the function is given by \( f(x) = 2x + 2 \). Since this function applies to all values greater than 4, there is no open circle at any point in this range because the function is defined there.
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Condition for \( x \leq 2 \): For \( x \) values less than or equal to 2, the function is given by \( f(x) = \frac{1}{2}x + 1 \). Since this is defined for \( x \) values less than or equal to 2, there will be no open circle in this portion either, except at \( x = 2 \).
Now, we need to check what happens at \( x = 2 \):
- When \( x = 2 \), we can substitute it into the function defined for that interval: \[ f(2) = \frac{1}{2}(2) + 1 = 1 + 1 = 2 \]
- However, since the second piece of the function (i.e., \( 2x + 2 \)) does not include \( x = 2 \) (it is defined only for \( x > 4 \)), there will be an open circle at the point \( (2, 2) \).
Thus, the coordinate point with an open circle in the graph of this piecewise function is: \[ (2, 2) \]