Question
Graphthe piecewise function f(x)={x+5 if x≤−22x+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 .
Yes, at (−2,−1).
Yes, at left parenthesis negative 2 comma negative 1 right parenthesis .
No, all endpoints are closed.
No, all endpoints are closed.
Yes, at (−2,3).
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 .
Yes, at (−2,−1).
Yes, at left parenthesis negative 2 comma negative 1 right parenthesis .
No, all endpoints are closed.
No, all endpoints are closed.
Yes, at (−2,3).
Answers
GPT-4o mini
To analyze the piecewise function \( f(x) \):
\[
f(x) =
\begin{cases}
x + 5 & \text{if } x \leq -2 \\
2x + 3 & \text{if } x > -2
\end{cases}
\]
### 1. Finding the y-values at the endpoint \( x = -2 \):
- For \( x \leq -2 \) (specifically at \( x = -2 \)):
\[
f(-2) = -2 + 5 = 3
\]
- For \( x > -2 \) (specifically at \( x = -2 \)):
\[
\text{This piece of the function does not include } x = -2.
\]
However, we can find \( f(x) \) approaching \( -2 \) from the right:
\[
\lim_{x \to -2^+} f(x) = 2(-2) + 3 = -4 + 3 = -1
\]
### 2. Analyzing the endpoints:
- At \( x = -2\):
- The function value is \( 3 \) (from the left side).
- The limit from the right approaches \( -1 \).
This means that the graph will have an open endpoint at the point \( (-2, -1) \), because the value at \( x = -2 \) does not include \( -2 \) in the second piece of the piecewise function (which only includes values greater than \( -2 \)).
### Conclusion:
**Answer**: Yes, at \( (-2, 3) \) and \( (-2, -1) \).
### Final Response:
Yes, at (−2,3) and (−2,−1).
\[
f(x) =
\begin{cases}
x + 5 & \text{if } x \leq -2 \\
2x + 3 & \text{if } x > -2
\end{cases}
\]
### 1. Finding the y-values at the endpoint \( x = -2 \):
- For \( x \leq -2 \) (specifically at \( x = -2 \)):
\[
f(-2) = -2 + 5 = 3
\]
- For \( x > -2 \) (specifically at \( x = -2 \)):
\[
\text{This piece of the function does not include } x = -2.
\]
However, we can find \( f(x) \) approaching \( -2 \) from the right:
\[
\lim_{x \to -2^+} f(x) = 2(-2) + 3 = -4 + 3 = -1
\]
### 2. Analyzing the endpoints:
- At \( x = -2\):
- The function value is \( 3 \) (from the left side).
- The limit from the right approaches \( -1 \).
This means that the graph will have an open endpoint at the point \( (-2, -1) \), because the value at \( x = -2 \) does not include \( -2 \) in the second piece of the piecewise function (which only includes values greater than \( -2 \)).
### Conclusion:
**Answer**: Yes, at \( (-2, 3) \) and \( (-2, -1) \).
### Final Response:
Yes, at (−2,3) and (−2,−1).