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)
left parenthesis 1 comma negative 2 right parenthesis

(1,1)
left parenthesis 1 comma 1 right parenthesis

(−1,−2)
left parenthesis negative 1 comma negative 2 right parenthesis

(−1,1)

Answer 1

To determine which point should have an open circle in the graph of the piecewise function \( f(x) =
\begin{cases}
-x - 1 & \text{if } x < 1 \\
-x + 2 & \text{if } x \geq 1
\end{cases} \), we need to look at the behavior of the function at \( x = 1 \).

1. For \( x < 1 \): The function is defined as \( f(x) = -x - 1 \). At \( x = 1 \), we can calculate \( f(1) \) using this piece:
\[
f(1) = -1 - 1 = -2.
\]
However, this piece is not valid at \( x = 1 \), because the condition requires \( x \) to be strictly less than 1.

2. For \( x \geq 1 \): The function is defined as \( f(x) = -x + 2 \). We can calculate this at \( x = 1 \) as well:
\[
f(1) = -1 + 2 = 1.
\]
Here, this piece is valid at \( x = 1 \).

Since the function value for the piece where \( x < 1 \) does not apply at \( x = 1 \), we will place an **open circle** at the point \( (1, -2) \) indicating that this value is not included in the graph.

To summarize, the point that should have an open circle is \( (1, -2) \).