Chapter 1: Problem 6
Find the domain and range of each function. $$G(t)=\frac{2}{t^{2}-16}$$
Short Answer
Expert verified
Domain: \((-\infty, -4) \cup (-4, 4) \cup (4, \infty)\); Range: \((-\infty, 0) \cup (0, \infty)\).
Step by step solution
01
Identify Domain Restrictions
The domain of a function is all the possible input values (in this case, all possible values of \( t \)) for which the function is defined. For the function \( G(t)=\frac{2}{t^{2}-16} \), the denominator \( t^2 - 16 \) cannot be zero, otherwise the function would be undefined. Set the denominator equal to zero and solve for \( t \):\[t^2 - 16 = 0\]This factors to \((t - 4)(t + 4) = 0\), giving that \( t = 4 \) or \( t = -4 \). Thus, \( t \) cannot be 4 or -4.
02
Define the Domain
With \( t \) not being able to be 4 or -4, the domain of \( G(t) \) includes all real numbers except 4 and -4. Therefore, the domain is \[ \{ t \in \mathbb{R} : t eq 4, t eq -4 \} \text{ or } (-\infty, -4) \cup (-4, 4) \cup (4, \infty) \].
03
Identify Range Restrictions
The range of the function involves identifying all possible output values \( G(t) \). The only restriction that comes from the function \( G(t)=\frac{2}{t^{2}-16} \) is that as \( t \) approaches 4 or -4, \( G(t) \) can approach any real number except for particular values that mirror the undefined or asymptotic behaviors. However, since \( t^2 - 16 \) can result in any non-zero number, the entire function \( G(t) \) outputs non-zero values.
04
Define the Range
Given that \( t \) approaching 4 or -4 results in a vertical asymptote, the output is undefined at those points. However, for \( t \) away from these values, \( G(t) \) can output any real number except zero, because it effectively creates a vertical asymptote with continuous output values on either side of zero. Therefore, the range is \[ G(t) eq 0 \] or \( (-\infty, 0) \cup (0, \infty) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function Analysis
Function analysis helps us understand the behavior and properties of a mathematical function. When analyzing functions such as \( G(t) = \frac{2}{t^2 - 16} \), we focus on identifying restrictions and key characteristics. One primary aspect is the **domain**, which represents all possible inputs. For \( G(t) \), we determine that the function is undefined when the denominator equals zero. By solving \( t^2 - 16 = 0 \), we find the function is undefined at \( t = 4 \) and \( t = -4 \). This indicates that the domain excludes these values:
- Domain: \( (-\infty, -4) \cup (-4, 4) \cup (4, \infty) \)
- Range: \( (-\infty, 0) \cup (0, \infty) \)
Asymptotes
Asymptotes play a crucial role in understanding functions, especially when analyzing rational functions like \( G(t) = \frac{2}{t^2 - 16} \). An **asymptote** is a line that a function approaches but does not touch or cross. In this function, vertical asymptotes occur where the function is undefined—at \( t = 4 \) and \( t = -4 \). These are identified by setting the denominator \( t^2 - 16 \) to zero. The function nears infinity or negative infinity close to these points, making
- Vertical asymptotes: \( t = 4 \) and \( t = -4 \)
- Horizontal asymptote: \( y = 0 \)
Rational Functions
Rational functions are expressions of the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials, typically with \( Q(x) \) not equal to zero. The function \( G(t) = \frac{2}{t^2 - 16} \) is a classic example, where the numerator \( P(t) \) is 2 and the denominator \( Q(t) \) is \( t^2 - 16 \). This function type often features:
- **Domain restrictions**: Determined by the values that make the denominator zero, hence the exclusions \( t eq 4 \) and \( t eq -4 \).
- **Discontinuities**: Places where the function is not continuous due to undefined points.
- **End behavior**: Driven by the function's degree and leads to asymptotic behavior, often described by horizontal and oblique asymptotes.