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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 answer

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.