To determine which coordinate point will have an open circle in the given piecewise function:
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Let's analyze the defined pieces of the function:
- For \( f(x) = 2x + 2 \) when \( x > 4 \).
- For \( f(x) = \frac{1}{2}x + 1 \) when \( x \leq 2 \).
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The point to investigate based on your hint is \( (4, ?) \). Since the first piece of the function only applies for \( x > 4 \), we need to check the value of \( f(x) \) at \( x = 4 \) to determine whether there is an open circle.
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Let's evaluate:
- The function \( f(x) = 2x + 2 \) is valid for \( x > 4 \). As \( x \) approaches 4 from the right, the value is: \[ f(4) = 2(4) + 2 = 8 + 2 = 10. \] However, this is the limit from the right side and is not the actual value of the function at \( x = 4 \).
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The piece \( f(x) = \frac{1}{2}x + 1 \) does not cover \( x = 4 \) as it only applies for \( x \leq 2 \). Thus, at \( x = 4\), the function is not defined, meaning the function does not take a value at this specific point from either piece of the function.
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Therefore, since the function approaches the value 10 as the input approaches 4 from the right, but does not include it, we can conclude that there will be an open circle at the point:
\[ (4, 10). \]
In conclusion, the coordinate point \( (4, 10) \) will have an open circle on the graph of the piecewise function.