To determine which of the given piecewise relations defines a function, let's analyze each case.
Relation 1:
[tex]\[y = \left\{
\begin{array}{cl}
x^2, & x < -2 \\
0, & -2 \leq x \leq 4 \\
-x^2, & x \geq 4
\end{array}
\right.\][/tex]
We need to check the values at the boundaries of each interval:
- For [tex]\(x = -2\)[/tex]:
- In the first interval: [tex]\(x < -2 \Rightarrow x^2\)[/tex]
- In the second interval: [tex]\(-2 \leq x \leq 4 \Rightarrow 0\)[/tex]
Clearly, at [tex]\(x = -2\)[/tex], [tex]\(y\)[/tex] takes the value [tex]\(0\)[/tex]. This is consistent.
- For [tex]\(x = 4\)[/tex]:
- In the second interval: [tex]\(-2 \leq x \leq 4 \Rightarrow 0\)[/tex]
- In the third interval: [tex]\(x \geq 4 \Rightarrow -x^2\)[/tex]
At [tex]\(x = 4\)[/tex], the value of [tex]\(y\)[/tex] should be:
- [tex]\(0\)[/tex] from the second interval,
- [tex]\(-16\)[/tex] from the third interval.
Since [tex]\(y\)[/tex] takes different values [tex]\(0\)[/tex] and [tex]\(-16\)[/tex] at [tex]\(x = 4\)[/tex], this relation is not a function.
Relation 2:
[tex]\[y = \left\{
\begin{array}{cl}
x^2, & x \leq -2 \\
4, & -2 < x \leq 2 \\
x^2 + 1, & x \geq 2
\end{array}
\right.\][/tex]
Check the values at the boundaries of each interval:
- For [tex]\(x = -2\)[/tex]:
- In the first interval: [tex]\(x \leq -2 \Rightarrow x^2\)[/tex]
- In the second interval: [tex]\(-2 < x \leq 2 \Rightarrow 4\)[/tex]
At [tex]\(x = -2\)[/tex], the value of [tex]\(y\)[/tex] should be:
- [tex]\(4\)[/tex] from the second interval,
- [tex]\((-2)^2 = 4\)[/tex] from the first interval.
So, [tex]\(y = 4\)[/tex] at [tex]\(x = -2\)[/tex] in both cases. This is consistent.
- For [tex]\(x = 2\)[/tex]:
- In the second interval: [tex]\(-2 < x \leq 2 \Rightarrow 4\)[/tex]
- In the third interval: [tex]\(x \geq 2 \Rightarrow x^2 + 1\)[/tex]
At [tex]\(x = 2\)[/tex], the value of [tex]\(y\)[/tex] should be:
- [tex]\(4\)[/tex] from the second interval,
- [tex]\(2^2 + 1 = 5\)[/tex] from the third interval.
Since [tex]\(y\)[/tex] takes different values [tex]\(4\)[/tex] and [tex]\(5\)[/tex] at [tex]\(x = 2\)[/tex], this relation is not a function.
Relation 3:
[tex]\[y = \left\{
\begin{aligned}
-3x, & \quad x < -2 \\
3, & \quad 0 \leq x < 4 \\
2x, & \quad x \geq 4
\end{aligned}
\right.\][/tex]
Check the boundaries of each interval:
- For [tex]\(x = -2\)[/tex]:
- In the first interval: [tex]\(x < -2 \Rightarrow -3x\)[/tex]
Since [tex]\(-2\)[/tex] is not included in any interval, there is no issue at [tex]\(-2\)[/tex].
- For [tex]\(x = 0\)[/tex]:
- Not included in [tex]\(x < -2\)[/tex]
- In the second interval: [tex]\(0 \leq x < 4 \Rightarrow 3\)[/tex]
- Not included in [tex]\(x \geq 4\)[/tex]
Since [tex]\(0\)[/tex] is only included once, [tex]\(y\)[/tex] takes a unique value [tex]\(3\)[/tex], which is consistent.
- For [tex]\(x = 4\)[/tex]:
- Not included in [tex]\(x < -2\)[/tex]
- In the second interval: [tex]\(0 \leq x < 4 \Rightarrow 3\)[/tex]
- In the third interval: [tex]\(x \geq 4 \Rightarrow 2x\)[/tex]
At [tex]\(x = 4\)[/tex], the value of [tex]\(y\)[/tex] should be:
- [tex]\(3\)[/tex] from the second interval,
- [tex]\(8\)[/tex] from the third interval.
Since [tex]\(y\)[/tex] takes different values [tex]\(3\)[/tex] and [tex]\(8\)[/tex], this relation is not a function.
Relation 4:
[tex]\[y = \left\{
\begin{aligned}
-3x, & \quad x \leq -4 \\
3, & \quad -5 < x < 1
\end{aligned}
\right.\][/tex]
Check the boundaries of each interval:
- For [tex]\(x = -4\)[/tex]:
- In the first interval: [tex]\(x \leq -4 \Rightarrow -3x\)[/tex]
- In the second interval: [tex]\(-5 < x < 1 \Rightarrow 3\)[/tex]
At [tex]\(x = -4\)[/tex], the value of [tex]\(y\)[/tex] should be:
- [tex]\( -3(-4) = 12\)[/tex]
- Not included in the second interval.
Thus, this relation defines two different values at [tex]\(x = -4\)[/tex] and is not a function.
Since none of these piecewise relations fit the criteria to define a consistent function, the final conclusion is that none of them define a function.
