To determine if the endpoints for the given piecewise function are open or closed, we need to analyze the definitions and the conditions of each piece of the function. The function is defined as follows:
\[ f(x) = \begin{cases} 2x & \text{if } x < 2 \ 3x & \text{if } 2 \leq x < 4 \ 4x & \text{if } x > 5 \end{cases} \]
-
First Piece: \(2x\) for \(x < 2\)
- The endpoint at \(x = 2\) is not included (since it says \(x < 2\)), so it is open.
-
Second Piece: \(3x\) for \(2 \leq x < 4\)
- The endpoint at \(x = 2\) is included (since it says \(2 \leq x\)), while the endpoint at \(x = 4\) is not included (since it says \(x < 4\)). Therefore, \(x = 2\) is closed and \(x = 4\) is open.
-
Third Piece: \(4x\) for \(x > 5\)
- The endpoint at \(x = 5\) is not included (since it says \(x > 5\)), so it is open.
Now, to find the last point graphed on the second piece of the function (i.e., \(3x\)), we look at the interval defined for this piece, which is \(2 \leq x < 4\).
- The last point on this segment is at \(x = 4\), but since \(x = 4\) is not included (\(x < 4\)), it will not actually be part of the graph. Instead, the last point graphed when \(x\) approaches 4 from the left side is \(3(4 - \epsilon)\) for very small \(\epsilon > 0\).
Thus, the last point graphed for the second piece of the function is at \(x = 4\), but it is not included, meaning the value of \(f(4)\) is not defined in this piece.
1 answer