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)
left parenthesis 1 comma 1 right parenthesis

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 check the value of the function at the point where the criteria change, which is at \(x = 1\).

1. For \( x < 1 \):
\[
f(1) \text{ is not calculated with this part as } 1 \text{ is not less than } 1.
\]
However, we can evaluate the limit from the left:
\[
\lim_{x \to 1^-} f(x) = -1 - 1 = -2.
\]

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

Now, we see that:
- As \(x\) approaches 1 from the left, \(f(x) \to -2\).
- At \(x = 1\), \(f(1) = 1\).

Since the value of \(f(x)\) at \(x = 1\) is 1, but the limit from the left (as \(x\) approaches 1) is -2, there will be an open circle at \( (1, -2) \) because this point is not included in the function's output at \(x = 1\).

Thus, the point that should have an open circle is:

**(1, -2)**.