Pregunta
Decide whether each relation defines \( y \) as a function of \( x \). Give the domain and range. \( y=\frac{3}{x-3} \) Does this relation give a function? \( \begin{array}{l}\text { Yes } \\ \text { No } \\ \text { What is the domain? } \\ \text { (Type your answer in interval notation.) } \\ \text { What is the range? } \\ \text { (Type your answer in interval notation.) }\end{array} \) (1)
Ask by Campos Bartlett. in the United States
Feb 13,2025
Solución de tutoría real
Respuesta verificada por el tutor
Responder
Yes, the relation defines y as a function of x. The domain is all real numbers except x = 3, written as (–∞, 3) ∪ (3, ∞). The range is all real numbers except y = 0, written as (–∞, 0) ∪ (0, ∞).
Solución
The given relation is y = 3/(x – 3).
1. Does this relation give a function?
Yes, because for every allowable value of x, there is exactly one corresponding y.
2. What is the domain?
The expression is undefined when x – 3 = 0, i.e., x = 3.
Domain (in interval notation): (–∞, 3) ∪ (3, ∞)
3. What is the range?
Since the numerator (3) is never 0, the fraction can never equal 0.
Range (in interval notation): (–∞, 0) ∪ (0, ∞)
Revisado y aprobado por el equipo de tutoría de UpStudy
Como
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Extra Insights
Yes, this relation does define \( y \) as a function of \( x \) because for every input \( x \) (except \( x=3 \), where the function is undefined), there is exactly one output \( y \). The domain of the function is all real numbers except \( x=3 \), which can be expressed in interval notation as \( (-\infty, 3) \cup (3, \infty) \). The range is all real numbers except \( y=0 \), since \( y \) approaches zero but never actually reaches it as \( x \) approaches positive or negative infinity, resulting in the range \( (-\infty, 0) \cup (0, \infty) \).
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