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
(1,1)
open paren 1 comma 1 close paren - no response given
(−1,1)
open paren negative 1 comma 1 close paren - no response given
(1,−2)
open paren 1 comma negative 2 close paren - no response given
(−1,−2)
open paren negative 1 comma negative 2 close paren - incorrect
Incorrect
?(1 point)
Responses
(1,1)
open paren 1 comma 1 close paren - no response given
(−1,1)
open paren negative 1 comma 1 close paren - no response given
(1,−2)
open paren 1 comma negative 2 close paren - no response given
(−1,−2)
open paren negative 1 comma negative 2 close paren - incorrect
Incorrect
Answers
GPT-4o mini
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 boundary point where the two cases switch, which is at \(x = 1\).
1. For \(x < 1\):
- We use the first piece: \(f(x) = -x - 1\).
- As we approach \(x = 1\) from the left (e.g., \(x = 0.9\)), we find \(f(0.9) = -0.9 - 1 = -1.9\).
2. For \(x \geq 1\):
- We use the second piece: \(f(x) = -x + 2\).
- At \(x = 1\), \(f(1) = -1 + 2 = 1\).
Now, at \(x = 1\):
- The limit from the left \( \lim_{x \to 1^-} f(x) = -1.9 \) (from the first case).
- The value from the right \(f(1) = 1\) (from the second case).
Since the function does not equal \(-1.9\) at \(x = 1\) when evaluating the left-hand limit, but equals \(1\) when evaluated at \(x = 1\), we should show that \(f(1)\) is defined but does not come from the left side.
So, we place an **open circle** at the point corresponding to the value of \(f\) from the first piece for \(x = 1\), which is \((1, -1.9)\).
Among the options given, the only point that corresponds correctly to an open circle at \(x = 1\) is \((1, -2)\), but that is not correct according to what we've evaluated. Instead, if the options were misinterpreted or mistyped in the context, the accurate point signifying the discontinuity at \(x=1\) would be \((1, f(1))\) showing it filled, and a point \( <1\) should have an open circle indicating where the graph is falling from on the left.
Therefore, none of the provided points directly correspond to a required open circle based on the given function division. But the correct representation for discontinuity at \(x=1\) leads to noting you would depict it at the bounds of \((1, -2)\) as the closest nature of possible representation might refer to \( f(x) \) values misplaced near or reflecting the \( -2 \).
Thus, the final identified point with an open circle is accurately at \((1, -2)\), while also being pointed out that \((1, 1)\) directly references the closed value.
\[
f(x) =
\begin{cases}
-x - 1 & \text{if } x < 1 \\
-x + 2 & \text{if } x \geq 1
\end{cases}
\]
we need to analyze the behavior of the function at the boundary point where the two cases switch, which is at \(x = 1\).
1. For \(x < 1\):
- We use the first piece: \(f(x) = -x - 1\).
- As we approach \(x = 1\) from the left (e.g., \(x = 0.9\)), we find \(f(0.9) = -0.9 - 1 = -1.9\).
2. For \(x \geq 1\):
- We use the second piece: \(f(x) = -x + 2\).
- At \(x = 1\), \(f(1) = -1 + 2 = 1\).
Now, at \(x = 1\):
- The limit from the left \( \lim_{x \to 1^-} f(x) = -1.9 \) (from the first case).
- The value from the right \(f(1) = 1\) (from the second case).
Since the function does not equal \(-1.9\) at \(x = 1\) when evaluating the left-hand limit, but equals \(1\) when evaluated at \(x = 1\), we should show that \(f(1)\) is defined but does not come from the left side.
So, we place an **open circle** at the point corresponding to the value of \(f\) from the first piece for \(x = 1\), which is \((1, -1.9)\).
Among the options given, the only point that corresponds correctly to an open circle at \(x = 1\) is \((1, -2)\), but that is not correct according to what we've evaluated. Instead, if the options were misinterpreted or mistyped in the context, the accurate point signifying the discontinuity at \(x=1\) would be \((1, f(1))\) showing it filled, and a point \( <1\) should have an open circle indicating where the graph is falling from on the left.
Therefore, none of the provided points directly correspond to a required open circle based on the given function division. But the correct representation for discontinuity at \(x=1\) leads to noting you would depict it at the bounds of \((1, -2)\) as the closest nature of possible representation might refer to \( f(x) \) values misplaced near or reflecting the \( -2 \).
Thus, the final identified point with an open circle is accurately at \((1, -2)\), while also being pointed out that \((1, 1)\) directly references the closed value.