Answer :
Alright, let's go through the given cases step by step to understand which values of the function [tex]\[f(x) = \lfloor x \rfloor - 1\][/tex] are equivalent.
### Definitions:
The floor function, denoted as [tex]\(\lfloor x \rfloor\)[/tex], returns the largest integer less than or equal to [tex]\(x\)[/tex].
### Calculations:
Let's break down the function [tex]\(f(x)\)[/tex] for each of the given values:
1. [tex]\(f(-1)\)[/tex] and [tex]\(f(-2)\)[/tex]:
[tex]\[ f(-1) = \lfloor -1 \rfloor - 1 = -1 - 1 = -2 \][/tex]
[tex]\[ f(-2) = \lfloor -2 \rfloor - 1 = -2 - 1 = -3 \][/tex]
Clearly, [tex]\(-2 \neq -3\)[/tex]. Therefore, [tex]\(f(-1)\)[/tex] and [tex]\(f(-2)\)[/tex] are not equivalent.
2. [tex]\(f(1)\)[/tex] and [tex]\(f(0)\)[/tex]:
[tex]\[ f(1) = \lfloor 1 \rfloor - 1 = 1 - 1 = 0 \][/tex]
[tex]\[ f(0) = \lfloor 0 \rfloor - 1 = 0 - 1 = -1 \][/tex]
Clearly, [tex]\(0 \neq -1\)[/tex]. Therefore, [tex]\(f(1)\)[/tex] and [tex]\(f(0)\)[/tex] are not equivalent.
3. [tex]\(f(-3)\)[/tex] and [tex]\(f(-2.1)\)[/tex]:
[tex]\[ f(-3) = \lfloor -3 \rfloor - 1 = -3 - 1 = -4 \][/tex]
[tex]\[ f(-2.1) = \lfloor -2.1 \rfloor - 1 = \lfloor -2.1 \rfloor - 1. Given \lfloor -2.1 \rfloor = -3, \text{ we get } f(-2.1) = -3 - 1 = -4 \][/tex]
Clearly, [tex]\(-4 = -4\)[/tex]. Therefore, [tex]\(f(-3)\)[/tex] and [tex]\(f(-2.1)\)[/tex] are equivalent.
4. [tex]\(f(2)\)[/tex] and [tex]\(f(1.9)\)[/tex]:
[tex]\[ f(2) = \lfloor 2 \rfloor - 1 = 2 - 1 = 1 \][/tex]
[tex]\[ f(1.9) = \lfloor 1.9 \rfloor - 1 = \lfloor 1.9 \rfloor - 1. Given \lfloor 1.9 \rfloor = 1, \text{ we get } f(1.9) = 1 - 1 = 0 \][/tex]
Clearly, [tex]\(1 \neq 0\)[/tex]. Therefore, [tex]\(f(2)\)[/tex] and [tex]\(f(1.9)\)[/tex] are not equivalent.
### Summary:
From these calculations, we find that the values of the function are equivalent in only one case:
- [tex]\(f(-3)\)[/tex] and [tex]\(f(-2.1)\)[/tex] are equivalent.
The other pairs ([tex]\(f(-1)\)[/tex] and [tex]\(f(-2)\)[/tex], [tex]\(f(1)\)[/tex] and [tex]\(f(0)\)[/tex], [tex]\(f(2)\)[/tex] and [tex]\(f(1.9)\)[/tex]) are not equivalent.
### Definitions:
The floor function, denoted as [tex]\(\lfloor x \rfloor\)[/tex], returns the largest integer less than or equal to [tex]\(x\)[/tex].
### Calculations:
Let's break down the function [tex]\(f(x)\)[/tex] for each of the given values:
1. [tex]\(f(-1)\)[/tex] and [tex]\(f(-2)\)[/tex]:
[tex]\[ f(-1) = \lfloor -1 \rfloor - 1 = -1 - 1 = -2 \][/tex]
[tex]\[ f(-2) = \lfloor -2 \rfloor - 1 = -2 - 1 = -3 \][/tex]
Clearly, [tex]\(-2 \neq -3\)[/tex]. Therefore, [tex]\(f(-1)\)[/tex] and [tex]\(f(-2)\)[/tex] are not equivalent.
2. [tex]\(f(1)\)[/tex] and [tex]\(f(0)\)[/tex]:
[tex]\[ f(1) = \lfloor 1 \rfloor - 1 = 1 - 1 = 0 \][/tex]
[tex]\[ f(0) = \lfloor 0 \rfloor - 1 = 0 - 1 = -1 \][/tex]
Clearly, [tex]\(0 \neq -1\)[/tex]. Therefore, [tex]\(f(1)\)[/tex] and [tex]\(f(0)\)[/tex] are not equivalent.
3. [tex]\(f(-3)\)[/tex] and [tex]\(f(-2.1)\)[/tex]:
[tex]\[ f(-3) = \lfloor -3 \rfloor - 1 = -3 - 1 = -4 \][/tex]
[tex]\[ f(-2.1) = \lfloor -2.1 \rfloor - 1 = \lfloor -2.1 \rfloor - 1. Given \lfloor -2.1 \rfloor = -3, \text{ we get } f(-2.1) = -3 - 1 = -4 \][/tex]
Clearly, [tex]\(-4 = -4\)[/tex]. Therefore, [tex]\(f(-3)\)[/tex] and [tex]\(f(-2.1)\)[/tex] are equivalent.
4. [tex]\(f(2)\)[/tex] and [tex]\(f(1.9)\)[/tex]:
[tex]\[ f(2) = \lfloor 2 \rfloor - 1 = 2 - 1 = 1 \][/tex]
[tex]\[ f(1.9) = \lfloor 1.9 \rfloor - 1 = \lfloor 1.9 \rfloor - 1. Given \lfloor 1.9 \rfloor = 1, \text{ we get } f(1.9) = 1 - 1 = 0 \][/tex]
Clearly, [tex]\(1 \neq 0\)[/tex]. Therefore, [tex]\(f(2)\)[/tex] and [tex]\(f(1.9)\)[/tex] are not equivalent.
### Summary:
From these calculations, we find that the values of the function are equivalent in only one case:
- [tex]\(f(-3)\)[/tex] and [tex]\(f(-2.1)\)[/tex] are equivalent.
The other pairs ([tex]\(f(-1)\)[/tex] and [tex]\(f(-2)\)[/tex], [tex]\(f(1)\)[/tex] and [tex]\(f(0)\)[/tex], [tex]\(f(2)\)[/tex] and [tex]\(f(1.9)\)[/tex]) are not equivalent.