Answer :
To determine which piecewise relations define a function, we need to verify that for every [tex]\( x \)[/tex] in the domain of the relation, there is exactly one corresponding [tex]\( y \)[/tex]. Let's evaluate each of the provided piecewise relations carefully.
### Relation 1:
[tex]\[ y = \begin{cases} x^2 & x < -2 \\ 0 & -2 \leq x \leq 4 \\ -x^2 & x \geq 4 \end{cases} \][/tex]
- For [tex]\( x < -2 \)[/tex], [tex]\( y = x^2 \)[/tex].
- For [tex]\(-2 \leq x \leq 4 \)[/tex], [tex]\( y = 0 \)[/tex].
- For [tex]\( x \geq 4 \)[/tex], [tex]\( y = -x^2 \)[/tex].
For each interval of [tex]\( x \)[/tex], there is a uniquely defined [tex]\( y \)[/tex]. Therefore, this piecewise relation defines a function.
### Relation 2:
[tex]\[ y = \begin{cases} x^2 & x \leq -2 \\ 4 & -2 < x \leq 2 \\ x^2 + 1 & x \geq 2 \end{cases} \][/tex]
- For [tex]\( x \leq -2 \)[/tex], [tex]\( y = x^2 \)[/tex].
- For [tex]\(-2 < x \leq 2 \)[/tex], [tex]\( y = 4 \)[/tex].
- For [tex]\( x \geq 2 \)[/tex], [tex]\( y = x^2 + 1 \)[/tex].
For each interval of [tex]\( x \)[/tex], there is a uniquely defined [tex]\( y \)[/tex]. Therefore, this piecewise relation defines a function.
### Relation 3:
[tex]\[ y = \begin{cases} -3x & x < -2 \\ 3 & 0 \leq x < 4 \\ 2x & x \geq 4 \end{cases} \][/tex]
- For [tex]\( x < -2 \)[/tex], [tex]\( y = -3x \)[/tex].
- For [tex]\( 0 \leq x < 4 \)[/tex], [tex]\( y = 3 \)[/tex].
- For [tex]\( x \geq 4 \)[/tex], [tex]\( y = 2x \)[/tex].
For each interval of [tex]\( x \)[/tex], there is a uniquely defined [tex]\( y \)[/tex]. Therefore, this piecewise relation defines a function.
### Relation 4:
[tex]\[ y = \begin{cases} -3x & x \leq -4 \\ 3 & -5 < x < 1 \end{cases} \][/tex]
- For [tex]\( x \leq -4 \)[/tex], [tex]\( y = -3x \)[/tex].
- For [tex]\( -5 < x < 1 \)[/tex], [tex]\( y = 3 \)[/tex].
For each interval of [tex]\( x \)[/tex], there is a uniquely defined [tex]\( y \)[/tex]. Therefore, this piecewise relation defines a function.
In conclusion, all the given piecewise relations define functions.
Therefore, the relations that define a function are:
1, 2, 3, and 4
### Relation 1:
[tex]\[ y = \begin{cases} x^2 & x < -2 \\ 0 & -2 \leq x \leq 4 \\ -x^2 & x \geq 4 \end{cases} \][/tex]
- For [tex]\( x < -2 \)[/tex], [tex]\( y = x^2 \)[/tex].
- For [tex]\(-2 \leq x \leq 4 \)[/tex], [tex]\( y = 0 \)[/tex].
- For [tex]\( x \geq 4 \)[/tex], [tex]\( y = -x^2 \)[/tex].
For each interval of [tex]\( x \)[/tex], there is a uniquely defined [tex]\( y \)[/tex]. Therefore, this piecewise relation defines a function.
### Relation 2:
[tex]\[ y = \begin{cases} x^2 & x \leq -2 \\ 4 & -2 < x \leq 2 \\ x^2 + 1 & x \geq 2 \end{cases} \][/tex]
- For [tex]\( x \leq -2 \)[/tex], [tex]\( y = x^2 \)[/tex].
- For [tex]\(-2 < x \leq 2 \)[/tex], [tex]\( y = 4 \)[/tex].
- For [tex]\( x \geq 2 \)[/tex], [tex]\( y = x^2 + 1 \)[/tex].
For each interval of [tex]\( x \)[/tex], there is a uniquely defined [tex]\( y \)[/tex]. Therefore, this piecewise relation defines a function.
### Relation 3:
[tex]\[ y = \begin{cases} -3x & x < -2 \\ 3 & 0 \leq x < 4 \\ 2x & x \geq 4 \end{cases} \][/tex]
- For [tex]\( x < -2 \)[/tex], [tex]\( y = -3x \)[/tex].
- For [tex]\( 0 \leq x < 4 \)[/tex], [tex]\( y = 3 \)[/tex].
- For [tex]\( x \geq 4 \)[/tex], [tex]\( y = 2x \)[/tex].
For each interval of [tex]\( x \)[/tex], there is a uniquely defined [tex]\( y \)[/tex]. Therefore, this piecewise relation defines a function.
### Relation 4:
[tex]\[ y = \begin{cases} -3x & x \leq -4 \\ 3 & -5 < x < 1 \end{cases} \][/tex]
- For [tex]\( x \leq -4 \)[/tex], [tex]\( y = -3x \)[/tex].
- For [tex]\( -5 < x < 1 \)[/tex], [tex]\( y = 3 \)[/tex].
For each interval of [tex]\( x \)[/tex], there is a uniquely defined [tex]\( y \)[/tex]. Therefore, this piecewise relation defines a function.
In conclusion, all the given piecewise relations define functions.
Therefore, the relations that define a function are:
1, 2, 3, and 4
