Asked by Kitty-
Graph the 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 .
Yes, at (−2,3).
Yes, at left parenthesis negative 2 comma 3 right parenthesis .
No, all endpoints are closed.
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 .
Yes, at (−2,3).
Yes, at left parenthesis negative 2 comma 3 right parenthesis .
No, all endpoints are closed.
Answers
Answered by
GPT-4o mini
To analyze the piecewise function \( f(x) \) defined as:
\[
f(x) =
\begin{cases}
x + 5 & \text{if } x \leq -2 \\
2x + 3 & \text{if } x > -2
\end{cases}
\]
we need to determine the behavior of the function at the boundary point \( x = -2 \).
1. **For \( x \leq -2 \):** Use the equation \( f(x) = x + 5 \).
- When \( x = -2 \):
\[
f(-2) = -2 + 5 = 3
\]
So, the function has a closed endpoint at the point \((-2, 3)\).
2. **For \( x > -2 \):** Use the equation \( f(x) = 2x + 3 \).
- As \( x \) approaches \(-2\) from the right (greater than -2):
\[
f(-2) = 2(-2) + 3 = -4 + 3 = -1
\]
Here, the function approaches \((-2, -1)\) but does not actually reach that point since the domain is restricted to \( x > -2\). Thus, there is an open endpoint at this point.
In conclusion, the graph will have:
- A closed endpoint at \((-2, 3)\) because \( x = -2 \) is included in the first piece of the function.
- An open endpoint at \((-2, -1)\) because \( x = -2 \) is not included in the second piece of the function.
The correct response is:
**Yes, at (−2, 3) and (−2, −1).**
\[
f(x) =
\begin{cases}
x + 5 & \text{if } x \leq -2 \\
2x + 3 & \text{if } x > -2
\end{cases}
\]
we need to determine the behavior of the function at the boundary point \( x = -2 \).
1. **For \( x \leq -2 \):** Use the equation \( f(x) = x + 5 \).
- When \( x = -2 \):
\[
f(-2) = -2 + 5 = 3
\]
So, the function has a closed endpoint at the point \((-2, 3)\).
2. **For \( x > -2 \):** Use the equation \( f(x) = 2x + 3 \).
- As \( x \) approaches \(-2\) from the right (greater than -2):
\[
f(-2) = 2(-2) + 3 = -4 + 3 = -1
\]
Here, the function approaches \((-2, -1)\) but does not actually reach that point since the domain is restricted to \( x > -2\). Thus, there is an open endpoint at this point.
In conclusion, the graph will have:
- A closed endpoint at \((-2, 3)\) because \( x = -2 \) is included in the first piece of the function.
- An open endpoint at \((-2, -1)\) because \( x = -2 \) is not included in the second piece of the function.
The correct response is:
**Yes, at (−2, 3) and (−2, −1).**
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