To analyze the piecewise function \( f(x) \), we first need to define it according to the conditions provided:
- \( f(x) = x + 5 \) if \( x \leq -2 \)
- \( f(x) = 2x + 3 \) if \( x > -2 \)
Next, let's determine the values of \( f(x) \) at the endpoint \( x = -2 \):
- For \( x = -2 \):
- Using the first piece: \( f(-2) = -2 + 5 = 3 \)
According to the definition of the piecewise function, since \( -2 \) is included in the first interval, we plot the point \((-2, 3)\) as a closed endpoint.
Now, we find the limit of \( f(x) \) from the right side of \( -2\) (using the second piece):
- For \( x = -2 \) approaching from the right (i.e., \( x > -2 \)):
- Using the second piece: \( f(-2) \) would not be defined as the domain does not include \( -2 \) for this piece, so it is actually evaluated as \( 2(-2) + 3 = -4 + 3 = -1\).
Thus, at \( x = -2 \):
- The left-side limit (from the first piece) yields \( 3 \).
- The right-side limit (from the second piece) yields \( -1 \).
Since the limit as \( x \) approaches \(-2\) from the right does not equal the value from the left, the point \((-2, -1)\) will be an open endpoint because \( x = -2 \) is not included in the second piece.
To summarize:
- At the point \( (-2, 3) \), the endpoint is closed because the function is defined there under the condition \( x \leq -2 \).
- At the point \( (-2, -1) \), the endpoint is open because that value corresponds to the limit approaching from the right (but is not included).
Conclusion about endpoints:
- Yes, the graph will have open endpoints at: \((-2, -1)\)
- Yes, the graph will have a closed endpoint at: \((-2, 3)\)
In answer to your question, the correct selections are:
- Yes, at \((-2, -1)\) (open)
- Yes, at \((-2, 3)\) (closed)
- All other endpoints are closed is incorrect because there is an open endpoint at \((-2, -1)\).