Question

Which point should have an open circle in the graph of the piecewise function f(x)={−x−1 if x<1−x+2 if x≥1
?(1 point)
Responses

(1,−2)
open paren 1 comma negative 2 close paren - no response given

(−1,1)
open paren negative 1 comma 1 close paren - no response given

(−1,−2)
open paren negative 1 comma negative 2 close paren - incorrect

(1,1)

Answer 1

To determine which point should have an open circle in the graph of the piecewise function \( f(x) \), we first analyze the function:

\[
f(x) = \begin{cases}
-x - 1 & \text{if } x < 1 \\
-x + 2 & \text{if } x \geq 1
\end{cases}
\]

The piecewise function has different definitions for \( x < 1 \) and \( x \geq 1 \).

1. For \( x < 1 \), the function is \( f(x) = -x - 1 \).
2. For \( x \geq 1 \), the function is \( f(x) = -x + 2 \).

Now, let's evaluate the function at the point where the two segments meet, which is \( x = 1 \):

- Using the first piece, where \( x < 1 \):
\[
f(1) \text{ is not defined in this segment.}
\]

- Using the second piece, where \( x \geq 1 \):
\[
f(1) = -1 + 2 = 1.
\]

At \( x = 1 \), the function has a defined value of \( 1 \) from the second piece, but it's not defined from the first piece. This implies that the point \( (1, 1) \) is included in the graph, but since the function is not defined for \( x = 1 \) from the first piece, we must indicate this with an open circle at that point.

Therefore, the point that should have an open circle in the graph is **(1, 1)**.