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 negative 1 comma negative 2 right parenthesis

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

(−1,1)
left parenthesis negative 1 comma 1 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 evaluate the function at the boundary point \( x = 1 \).

1. For \( x < 1 \):
\[
f(x) = -x - 1
\]
When we approach \( x = 1 \) from the left (\( x \to 1^- \)):
\[
f(1^-) = -1 - 1 = -2
\]

2. For \( x \geq 1 \):
\[
f(x) = -x + 2
\]
When we evaluate at \( x = 1 \):
\[
f(1) = -1 + 2 = 1
\]

Thus, at \( x = 1 \), the function takes the value of 1, but as \( x \) approaches 1 from the left, the function approaches -2. Therefore, at the point \( (1, -2) \), there will be an open circle because \( f(1) \) does not equal \(-2\), but the left-hand limit does.

Given this analysis, the point that should have an open circle is \( (1, -2) \).