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

Answer 1

In the given 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 points given, particularly how they relate to the transition at \( x = 1 \).

1. For \( x < 1 \), the function is defined as \( f(x) = -x - 1 \).
- At \( x = -1 \):
\[
f(-1) = -(-1) - 1 = 0 - 1 = -1
\]
This point is \((-1, -1)\), but it does not give a direct implication for an open circle.

2. For \( x \geq 1 \), the function is \( f(x) = -x + 2 \).
- At \( x = 1 \):
\[
f(1) = -1 + 2 = 1
\]
This means we have point \( (1, 1) \) for \( x = 1 \).

Now let's check the points you've listed:

- \( (-1, 1) \) is not corresponding to any of the calculated values.
- \( (1, -2) \): This point does not match since \( f(1) = 1 \).
- \( (-1, -2) \): This point is unrelated since it's not a calculated value of the function.
- \( (1, 1) \): This point matches with the calculation \( f(1) = 1 \).

To summarize: at \( x = 1 \), \( f(x) = 1\) is included in the second piece of the piecewise function \( (-x + 2) \) since it specifies \( x \geq 1\).

However, at \( x = 1 \), we have a transition from the first function segment where \( f(1) \) would be calculated as \(-1 - 1 = -2\), but since \( f(1) \) is defined in the second segment, we denote this as having an open circle at \( (1, -2) \).

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

**(1, -2)**