To analyze the piecewise function \( f(x) \):
\[ f(x) = \begin{cases} x + 5 & \text{if } x \leq -2 \ 2x + 3 & \text{if } x > -2 \end{cases} \]
-
For the first piece \( f(x) = x + 5 \) when \( x \leq -2 \):
- At \( x = -2 \): \( f(-2) = -2 + 5 = 3 \) This point is included in this part of the function because the condition is \( x \leq -2 \).
-
For the second piece \( f(x) = 2x + 3 \) when \( x > -2 \):
- The domain starts just above -2. Therefore, it does not include \( x = -2 \).
- As \( x \) approaches -2 from the right, \( f(-2^+) = 2(-2) + 3 = -1 \). This means at \( x = -2 \), the function approaches the value -1, but this point is not included in the graph.
Now, let's summarize the behavior at \( x = -2 \):
- The function has a value of \( 3 \) when approaching from the left (included, closed endpoint) and a value of \( -1 \) when approaching from the right (not included, open endpoint).
Thus, the graph will have an open endpoint at \( (-2, -1) \).
Based on the responses given, the correct answer is:
Yes, at \((-2, 3)\) and \((-2, -1)\).
However, since \((-2, 3)\) is the only closed endpoint, the other is open, so the more accurate response would be:
Yes, at (-2, -1).
1 answer