Question
Graph the piecewise function f(x)={x+5 if xβ€β22x+3 if x>β2
π
(
π₯
)
=
{
π₯
+
5
if
π₯
β€
β
2
2
π₯
+
3
if
π₯
>
β
2
. Will the graph have any open endpoints? If yes, where?(1 point)
Responses
Yes, at (β2,3)
(
β
2
,
3
)
and (β2,β1)
(
β
2
,
β
1
)
.
Yes, at left parenthesis negative 2 comma 3 right parenthesis and left parenthesis negative 2 comma negative 1 right parenthesis .
Yes, at (β2,β1)
(
β
2
,
β
1
)
.
Yes, at left parenthesis negative 2 comma negative 1 right parenthesis .
No, all endpoints are closed.
No, all endpoints are closed.
Yes, at (β2,3)
(
β
2
,
3
)
.
π
(
π₯
)
=
{
π₯
+
5
if
π₯
β€
β
2
2
π₯
+
3
if
π₯
>
β
2
. Will the graph have any open endpoints? If yes, where?(1 point)
Responses
Yes, at (β2,3)
(
β
2
,
3
)
and (β2,β1)
(
β
2
,
β
1
)
.
Yes, at left parenthesis negative 2 comma 3 right parenthesis and left parenthesis negative 2 comma negative 1 right parenthesis .
Yes, at (β2,β1)
(
β
2
,
β
1
)
.
Yes, at left parenthesis negative 2 comma negative 1 right parenthesis .
No, all endpoints are closed.
No, all endpoints are closed.
Yes, at (β2,3)
(
β
2
,
3
)
.
Answers
GPT-4o mini
To analyze the piecewise function given by:
\[
f(x) =
\begin{cases}
x + 5 & \text{if } x \leq -2 \\
2x + 3 & \text{if } x > -2
\end{cases}
\]
let's evaluate the function at \( x = -2 \):
1. For \( x \leq -2 \):
\[
f(-2) = -2 + 5 = 3
\]
2. For \( x > -2 \), we cannot directly plug in \( x = -2 \) because this part of the piecewise function does not include \( -2 \).
Since the function is defined at \( x = -2 \) as \( 3 \) and the second piece is not defined at \( x = -2 \) (it only applies to \( x > -2 \)), the point \( (-2, 3) \) is included in the graph. However, the value \( f(-2) \) from the second part of the function is not included. Therefore, we can infer that the endpoint at \( x = -2 \) has an open point on the line segment of the function defined by \( 2x + 3 \).
Thus, the graph will have an open endpoint at \((-2, 3)\).
### Final Answer:
Yes, at \((-2, 3)\).
\[
f(x) =
\begin{cases}
x + 5 & \text{if } x \leq -2 \\
2x + 3 & \text{if } x > -2
\end{cases}
\]
let's evaluate the function at \( x = -2 \):
1. For \( x \leq -2 \):
\[
f(-2) = -2 + 5 = 3
\]
2. For \( x > -2 \), we cannot directly plug in \( x = -2 \) because this part of the piecewise function does not include \( -2 \).
Since the function is defined at \( x = -2 \) as \( 3 \) and the second piece is not defined at \( x = -2 \) (it only applies to \( x > -2 \)), the point \( (-2, 3) \) is included in the graph. However, the value \( f(-2) \) from the second part of the function is not included. Therefore, we can infer that the endpoint at \( x = -2 \) has an open point on the line segment of the function defined by \( 2x + 3 \).
Thus, the graph will have an open endpoint at \((-2, 3)\).
### Final Answer:
Yes, at \((-2, 3)\).