Let's go through each part of the problem:
Part 1: Graphing the piecewise function
The function \( f(x) \) is defined as:
- \( f(x) = 6x - 9 \) for \( x \leq 3 \)
- \( f(x) = x^2 \) for \( x > 3 \)
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For \( x \leq 3 \):
- At \( x = 3 \): \( f(3) = 6(3) - 9 = 18 - 9 = 9 \)
- At \( x = 2 \): \( f(2) = 6(2) - 9 = 12 - 9 = 3 \)
- The line will have a slope of 6, starting from the point (3, 9) down to the left.
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For \( x > 3 \):
- At \( x = 3 \): \( f(3) \) is already calculated and equals 9. This point is open as the function switches to \( x^2 \) after \( x = 3 \).
- At \( x = 4 \): \( f(4) = 4^2 = 16 \)
The graph should:
- Include a closed point at (3, 9) and rise on the right side with the curve for \( x^2 \).
- Include an open point at (3, 9) for the linear function's left part.
Choosing the correct graph: Based on the description, the correct answer is likely B (where there is an open point at (3, 9) for the line and a closed point for the quadratic starting at that same vertical height).
Part 2: Evaluating the function.
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For \( f(1) \): Since \( 1 \leq 3 \), we use the first part of the function: \[ f(1) = 6(1) - 9 = 6 - 9 = -3 \]
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For \( f(5) \): Since \( 5 > 3 \), we use the second part of the function: \[ f(5) = 5^2 = 25 \]
Results:
- \( f(1) = -3 \)
- \( f(5) = 25 \)
Part 3: Domain of the function.
The function is defined for all \( x \) (both parts cover the cases for \( x \) less than or equal to 3 and greater than 3). Thus:
- The domain is all real numbers.
So the correct answer is: D. The domain of the function is all real numbers.
Summary:
- a. Choose graph B.
- b. \( f(1) = -3 \) and \( f(5) = 25 \).
- c. The domain of the function is all real numbers (D).