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, 1)
(-1, 1)
(-1, -2)
(1, -2)

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 behavior of the function at the point \( x = 1 \).

1. For \( x < 1 \), the function is defined as \( f(x) = -x - 1 \). To find \( f(1) \) with this part, we find the limit as \( x \) approaches 1 from the left:
\[
f(1^-) = -1 - 1 = -2.
\]

2. For \( x \geq 1 \), the function is defined as \( f(x) = -x + 2 \). Evaluating this at \( x = 1 \):
\[
f(1) = -1 + 2 = 1.
\]

From the above calculations, we find:
- The limit of \( f(x) \) as \( x \) approaches 1 from the left \( (f(1^-)) \) is \( -2 \).
- The value of \( f(1) \) is \( 1 \), which is determined by the second piece of the piecewise function.

Since \( f(1) \) differs from the limit approaching from the left, the point \( (1, -2) \) should be an open circle, while the dot at the point \( (1, 1) \) will be filled.

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