Question
Graph the piecewise function f(x)={x+5 if x≤−2 2x+3 if x>−2
. Will the graph have any open endpoints? If yes, where?(1 point)
Responses
Yes, at (−2,3)
and (−2,−1)
.
Yes, at left parenthesis negative 2 comma 3 right parenthesis and left parenthesis negative 2 comma negative 1 right parenthesis .
No, all endpoints are closed.
No, all endpoints are closed.
Yes, at (−2,−1)
.
Yes, at left parenthesis negative 2 comma negative 1 right parenthesis .
Yes, at (−2,3)
.
. Will the graph have any open endpoints? If yes, where?(1 point)
Responses
Yes, at (−2,3)
and (−2,−1)
.
Yes, at left parenthesis negative 2 comma 3 right parenthesis and left parenthesis negative 2 comma negative 1 right parenthesis .
No, all endpoints are closed.
No, all endpoints are closed.
Yes, at (−2,−1)
.
Yes, at left parenthesis negative 2 comma negative 1 right parenthesis .
Yes, at (−2,3)
.
Answers
GPT-4o mini
To analyze the piecewise function \( f(x) \) given by:
\[
f(x) =
\begin{cases}
x + 5 & \text{if } x \leq -2 \\
2x + 3 & \text{if } x > -2
\end{cases}
\]
1. **Finding the function at \( x = -2 \):**
- For \( x \leq -2 \):
\[
f(-2) = -2 + 5 = 3
\]
- For \( x > -2 \):
\[
\text{We do not need to evaluate } f(-2) \text{ here since it's defined for } x > -2.
\]
2. **Finding the limit as \( x \) approaches \(-2\) from the right:**
- For values just greater than \(-2\) (i.e., approaching \(-2\) from the right):
\[
f(-2^+) = 2(-2) + 3 = -4 + 3 = -1
\]
### Identifying open and closed endpoints:
- The value \( f(-2) = 3 \) exists and is defined for \( x = -2 \).
- However, as \( x \) approaches \(-2\) from the right, the function approaches \(-1\) (i.e., \( f(-2^+) = -1\)).
Thus:
- The graph of the piecewise function at \( x = -2 \) has a closed endpoint at \( (−2, 3) \) (since this value is included in the function for \( x \leq -2 \)).
- The function \( f(x) \) does not take the value \( -1 \) at \( x = -2 \), meaning this endpoint is not included on the graph. Therefore, there is an open endpoint at \( (−2, −1) \).
### Conclusion:
**Yes, there will be open endpoints.** The open endpoints are at:
- \( (-2, -1) \) is an open endpoint (since the value does not exist at that point).
- \( (-2, 3) \) is a closed endpoint (included in the function).
The correct responses to the question of whether the graph has open endpoints are:
**Yes, at (−2, −1).**
\[
f(x) =
\begin{cases}
x + 5 & \text{if } x \leq -2 \\
2x + 3 & \text{if } x > -2
\end{cases}
\]
1. **Finding the function at \( x = -2 \):**
- For \( x \leq -2 \):
\[
f(-2) = -2 + 5 = 3
\]
- For \( x > -2 \):
\[
\text{We do not need to evaluate } f(-2) \text{ here since it's defined for } x > -2.
\]
2. **Finding the limit as \( x \) approaches \(-2\) from the right:**
- For values just greater than \(-2\) (i.e., approaching \(-2\) from the right):
\[
f(-2^+) = 2(-2) + 3 = -4 + 3 = -1
\]
### Identifying open and closed endpoints:
- The value \( f(-2) = 3 \) exists and is defined for \( x = -2 \).
- However, as \( x \) approaches \(-2\) from the right, the function approaches \(-1\) (i.e., \( f(-2^+) = -1\)).
Thus:
- The graph of the piecewise function at \( x = -2 \) has a closed endpoint at \( (−2, 3) \) (since this value is included in the function for \( x \leq -2 \)).
- The function \( f(x) \) does not take the value \( -1 \) at \( x = -2 \), meaning this endpoint is not included on the graph. Therefore, there is an open endpoint at \( (−2, −1) \).
### Conclusion:
**Yes, there will be open endpoints.** The open endpoints are at:
- \( (-2, -1) \) is an open endpoint (since the value does not exist at that point).
- \( (-2, 3) \) is a closed endpoint (included in the function).
The correct responses to the question of whether the graph has open endpoints are:
**Yes, at (−2, −1).**