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 analyze the function at \(x = 1\).

1. **For \(x < 1\)**: The function is defined as \(f(x) = -x - 1\). If we substitute \(x = 1\) into this part, it doesn't apply because this part is only for \(x < 1\).
2. **For \(x = 1\)**: we look at the second part of the function, which is valid because it states \(x \geq 1\).
\[
f(1) = -1 + 2 = 1
\]

This means the point at \( (1, 1) \) exists in the graph. However, since \(x = 1\) is not included in the first part, we have to represent the value of \(f(1)\) but we will not include the point \( (1, -2) \) that would have come from the first piece of the function.

So the open circle represents that the point \( (1, -2) \) is not included but the point \( (1, 1)\) is included.

Thus, the correct answer is:

**(1, -2)** should have an open circle.