To determine which piecewise relations define functions, we need to ensure that each relation gives exactly one output \( y \) value for each input \( x \) value. Let's analyze each relation individually step-by-step:
### Relation 1
[tex]\[ y = \begin{cases}
x^2, & x < -2 \\
0, & -2 \leq x \leq 4 \\
-x^2, & x \geq 4
\end{cases} \][/tex]
1. For \( x < -2 \), \( y = x^2 \) has one value for each \( x \).
2. For \( -2 \leq x \leq 4 \), \( y = 0 \) is a single value for each \( x \).
3. For \( x \geq 4 \), \( y = -x^2 \) has one value for each \( x \).
Each segment of \( x \) provides a unique \( y \), which qualifies it as 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]
1. For \( x \leq -2 \), \( y = x^2 \) has one value for each \( x \).
2. For \( -2 < x \leq 2 \), \( y = 4 \) is a single value for each \( x \).
3. For \( x \geq 2 \), \( y = x^2 + 1 \) has one value for each \( x \).
Each segment of \( x \) provides a unique \( y \), which qualifies it as a function.
### Relation 3
[tex]\[ y = \begin{cases}
-3x, & x < -2 \\
3, & 0 \leq x < 4 \\
2x, & x \geq 4
\end{cases} \][/tex]
1. For \( x < -2 \), \( y = -3x \) has one value for each \( x \).
2. For \( 0 \leq x < 4 \), \( y = 3 \) is a single value for each \( x \).
3. For \( x \geq 4 \), \( y = 2x \) has one value for each \( x \).
Each segment of \( x \) provides a unique \( y \), which qualifies it as a function.
### Relation 4
[tex]\[ y = \begin{cases}
-3x, & x \leq -4 \\
3, & -5 < x < 1 \\
2x, & x \geq 1
\end{cases} \][/tex]
1. For \( x \leq -4 \), \( y = -3x \) has one value for each \( x \).
2. For \( -5 < x < 1 \), \( y = 3 \) is a single value for each \( x \).
3. For \( x \geq 1 \), \( y = 2x \) has one value for each \( x \).
Each segment of \( x \) provides a unique \( y \), which qualifies it as a function.
### Conclusion
All four piecewise relations:
1. \( y = \begin{cases} x^2, & x < -2 \\ 0, & -2 \leq x \leq 4 \\ -x^2, & x \geq 4 \end{cases} \)
2. \( y = \begin{cases} x^2, & x \leq -2 \\ 4, & -2 < x \leq 2 \\ x^2 + 1, & x \geq 2 \end{cases} \)
3. \( y = \begin{cases} -3x, & x < -2 \\ 3, & 0 \leq x < 4 \\ 2x, & x \geq 4 \end{cases} \)
4. \( y = \begin{cases} -3x, & x \leq -4 \\ 3, & -5 < x < 1 \\ 2x, & x \geq 1 \end{cases} \)
define a function, as each of them provides exactly one [tex]\( y \)[/tex] value for each [tex]\( x \)[/tex] value.
