To understand the endpoints of the piecewise function \( f(x) = \begin{cases} x + 5 & \text{if } x \leq -2 \ 2x + 3 & \text{if } x > -2 \end{cases} \), we can analyze each piece separately.
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For the first piece of the function, \( f(x) = x + 5 \) is applicable for \( x \leq -2 \). The point \( (-2, -2 + 5) = (-2, 3) \) is included because the piece applies to \( x \leq -2 \), meaning this point is included in the function.
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For the second piece, \( f(x) = 2x + 3 \) is applicable for \( x > -2 \). The limit as \( x \) approaches -2 from the right gives \( f(-2) = 2(-2) + 3 = -1 \), but this point is not included in this piece (since strictly \( x > -2 \)), indicating \( (-2, -1) \) is not part of the graph.
Hence, there is a discontinuity there:
- The endpoint at \( (-2, 3) \) is closed.
- The value \( (-2, -1) \) is not included, creating an open endpoint.
Therefore, the graph will have an open endpoint at (−2, −1).
Thus, the correct response is:
Yes, at (−2,−1).