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Which piecewise relation defines a function?

A. [tex]y=\left\{\begin{aligned} x^2, & \, x \ \textless \ -2 \\ 0, & \, -2 \leq x \leq 4 \\ -x^2, & \, x \geq 4 \end{aligned}\right.[/tex]

B. [tex]y=\left\{\begin{array}{cl} x^2, & \, x \leq -2 \\ 4, & \, -2 \ \textless \ x \leq 2 \\ x^2 + 1, & \, x \geq 2 \end{array}\right.[/tex]

C. [tex]y=\left\{\begin{aligned} -3x, & \, x \ \textless \ -2 \\ 3, & \, 0 \leq x \ \textless \ 4 \\ 2x, & \, x \geq 4 \end{aligned}\right.[/tex]

Sagot :

GinnyAnswer
To determine which piecewise relations define a function, we need to confirm that each piecewise relation assigns exactly one output value [tex]\( y \)[/tex] for each input value [tex]\( x \)[/tex]. Let's analyze each given piecewise relation in detail.

1. First Piecewise Relation:
[tex]\[ y = \begin{cases} x^2 & \text{if } x < -2 \\ 0 & \text{if } -2 \le x \le 4 \\ -x^2 & \text{if } x \ge 4 \end{cases} \][/tex]
- For [tex]\( x < -2 \)[/tex], [tex]\( y = x^2 \)[/tex], which is a single-valued function.
- For [tex]\( -2 \le x \le 4 \)[/tex], [tex]\( y = 0 \)[/tex], which is also a single-valued function.
- For [tex]\( x \ge 4 \)[/tex], [tex]\( y = -x^2 \)[/tex], which is again single-valued.

From this examination, each part of the piecewise relation assigns exactly one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex]. The first relation therefore defines a function.

2. Second Piecewise Relation:
[tex]\[ y = \begin{cases} x^2 & \text{if } x \le -2 \\ 4 & \text{if } -2 < x \le 2 \\ x^2 + 1 & \text{if } x \ge 2 \end{cases} \][/tex]
- For [tex]\( x \le -2 \)[/tex], [tex]\( y = x^2 \)[/tex], which is a single-valued function.
- For [tex]\( -2 < x \le 2 \)[/tex], [tex]\( y = 4 \)[/tex], which is also a single-valued function.
- For [tex]\( x \ge 2 \)[/tex], [tex]\( y = x^2 + 1 \)[/tex], which is again single-valued.

Thus, each part of this piecewise relation assigns exactly one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex]. The second relation also defines a function.

3. Third Piecewise Relation:
[tex]\[ y = \begin{cases} -3x & \text{if } x < -2 \\ 3 & \text{if } 0 \le x < 4 \\ 2x & \text{if } x \ge 4 \end{cases} \][/tex]
- For [tex]\( x < -2 \)[/tex], [tex]\( y = -3x \)[/tex], which is a single-valued function.
- For [tex]\( 0 \le x < 4 \)[/tex], [tex]\( y = 3 \)[/tex], which is also a single-valued function.
- For [tex]\( x \ge 4 \)[/tex], [tex]\( y = 2x \)[/tex], which is again single-valued.

However, this third relation does not cover the interval [tex]\( -2 \le x < 0 \)[/tex]. Specifically, for [tex]\( -2 \le x < 0 \)[/tex], there is no value defined for [tex]\( y \)[/tex]. Because of these gaps, this piecewise relation does not define a function over all real numbers.

Given this analysis, the piecewise relations that define a function are the first and second ones. Therefore, the correct options are [tex]\([1, 2]\)[/tex].
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