To analyze the piecewise function \( f(x) = \begin{cases} x + 5 & \text{if } x \leq -2 \ 2x + 3 & \text{if } x > -2 \end{cases} \), we need to evaluate the behavior of the function at the point \( x = -2 \).
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Check the first piece (when \( x \leq -2 \)):
- Substitute \( x = -2 \) into \( f(x) = x + 5 \): \[ f(-2) = -2 + 5 = 3 \] This means that at \( x = -2 \), the function value is 3, and this point will be included in the graph due to the \( \leq \) condition.
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Check the second piece (when \( x > -2 \)):
- The piece for \( f(x) = 2x + 3 \) does not include \( x = -2 \) because it only applies for \( x > -2 \). Thus, we check the limit as \( x \) approaches -2 from the right: \[ \lim_{x \to -2^+} f(x) = 2(-2) + 3 = -4 + 3 = -1 \] The function approaches -1 as \( x \) gets closer to -2 from the right side.
Summary of endpoints:
- At \( x = -2 \), the left piece gives \( f(-2) = 3 \) (included point).
- The right piece approaches \( -1 \), but does not include \( (-2, -1) \) since it only holds for \( x > -2 \).
Conclusion on open endpoints:
Since there is a different value from the left and the right when approaching \( x = -2 \), there is an open endpoint at \( (-2, -1) \), where the piece \( f(x) = 2x + 3 \) approaches but does not include.
Thus, the correct response is: Yes, at \( (-2, 3) \) and \( (-2, -1) \).
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