To determine whether the endpoints of the piecewise function
\[ f(x) = \begin{cases} 2x & \text{if } x < 2 \ 3x & \text{if } 2 \leq x < 4 \ 4x & \text{if } x > 5 \end{cases} \]
are open or closed, we need to analyze the conditions at the transition points of the piecewise function.
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For the first piece: \( f(x) = 2x \) for \( x < 2 \)
- The endpoint \( x = 2 \) is not included in this piece, which means it is open.
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For the second piece: \( f(x) = 3x \) for \( 2 \leq x < 4 \)
- Here, \( x = 2 \) is included (closed), and \( x = 4 \) is not included (open).
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For the third piece: \( f(x) = 4x \) for \( x > 5 \)
- The endpoint \( x = 5 \) is not included in this piece, meaning it is open.
Now, to answer the question about the last point graphed on the second piece of the function \( f(x) = 3x \):
- The last point of this piece would occur at \( x = 4 \). Since the condition for this piece is \( 2 \leq x < 4 \), the function does not include \( x = 4 \). Therefore, the last point graphed on this segment is at \( x = 4 \) (open).
In conclusion:
- At \( x = 2 \): Closed
- At \( x = 4 \): Open
- At \( x = 5 \): Open
The answer to your specific question is that the last point graphed on the second piece \( f(x) = 3x \) is at \( x = 4 \) (open).