Answer :
To determine where the open circle should be drawn for the given piecewise function
[tex]\[ f(x)=\begin{cases} -x, & x<0 \\ 1, & x \geq 0 \end{cases} \][/tex]
let's carefully analyze the definition of the function at [tex]\(x = 0\)[/tex].
1. For [tex]\(x < 0\)[/tex]:
The function is defined as [tex]\(f(x) = -x\)[/tex]. This means that for any negative value of [tex]\(x\)[/tex], [tex]\(f(x)\)[/tex] will be the negative of [tex]\(x\)[/tex].
2. For [tex]\(x \geq 0\)[/tex]:
The function is defined as [tex]\(f(x) = 1\)[/tex]. This means that for any value of [tex]\(x\)[/tex] equal to or greater than 0, the function value will be 1.
At [tex]\(x = 0\)[/tex], [tex]\(f(0) = 1\)[/tex].
3. Open Circle Consideration:
The point at which the function changes its definition needs careful consideration. When [tex]\(x = 0\)[/tex], the function value according to the [tex]\(x < 0\)[/tex] definition is [tex]\(f(0) = -0 = 0\)[/tex], but since this is not inclusive when [tex]\(x \geq 0\)[/tex], the function takes the value [tex]\(f(0) = 1\)[/tex].
Therefore, for the function [tex]\(f(x)\)[/tex] to be correctly represented:
- The transition occurs at [tex]\(x = 0\)[/tex], and the point [tex]\((0, 0)\)[/tex] should be represented as an open circle to indicate that the point [tex]\( (0, 0) \)[/tex] is not included in the domain for the [tex]\( -x \)[/tex] part and instead jumps to [tex]\( (0, 1) \)[/tex].
So, the open circle should be drawn at [tex]\((0, 0)\)[/tex].
Hence, the correct answer is:
[tex]\[ \boxed{(0, 0)} \][/tex]
[tex]\[ f(x)=\begin{cases} -x, & x<0 \\ 1, & x \geq 0 \end{cases} \][/tex]
let's carefully analyze the definition of the function at [tex]\(x = 0\)[/tex].
1. For [tex]\(x < 0\)[/tex]:
The function is defined as [tex]\(f(x) = -x\)[/tex]. This means that for any negative value of [tex]\(x\)[/tex], [tex]\(f(x)\)[/tex] will be the negative of [tex]\(x\)[/tex].
2. For [tex]\(x \geq 0\)[/tex]:
The function is defined as [tex]\(f(x) = 1\)[/tex]. This means that for any value of [tex]\(x\)[/tex] equal to or greater than 0, the function value will be 1.
At [tex]\(x = 0\)[/tex], [tex]\(f(0) = 1\)[/tex].
3. Open Circle Consideration:
The point at which the function changes its definition needs careful consideration. When [tex]\(x = 0\)[/tex], the function value according to the [tex]\(x < 0\)[/tex] definition is [tex]\(f(0) = -0 = 0\)[/tex], but since this is not inclusive when [tex]\(x \geq 0\)[/tex], the function takes the value [tex]\(f(0) = 1\)[/tex].
Therefore, for the function [tex]\(f(x)\)[/tex] to be correctly represented:
- The transition occurs at [tex]\(x = 0\)[/tex], and the point [tex]\((0, 0)\)[/tex] should be represented as an open circle to indicate that the point [tex]\( (0, 0) \)[/tex] is not included in the domain for the [tex]\( -x \)[/tex] part and instead jumps to [tex]\( (0, 1) \)[/tex].
So, the open circle should be drawn at [tex]\((0, 0)\)[/tex].
Hence, the correct answer is:
[tex]\[ \boxed{(0, 0)} \][/tex]
