Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![**Problem Statement:**
Determine the domain and range of the vector-valued function:
\[
\mathbf{r}(t) = \left\langle \ln(t-2), \frac{1}{t-4}, \sqrt{t} \right\rangle
\]
**Explanation:**
To find the domain of the vector-valued function \(\mathbf{r}(t)\), we need to determine the values of \(t\) for which each component of the function is defined.
1. **First Component \(\ln(t-2)\):**
- The logarithm function \(\ln(x)\) is defined for \(x > 0\).
- Thus, \(t-2 > 0 \Rightarrow t > 2\).
2. **Second Component \(\frac{1}{t-4}\):**
- The expression \(\frac{1}{x}\) is defined for \(x \neq 0\).
- Thus, \(t - 4 \neq 0 \Rightarrow t \neq 4\).
3. **Third Component \(\sqrt{t}\):**
- The square root function \(\sqrt{x}\) is defined for \(x \geq 0\).
- Thus, \(t \geq 0\).
Combining these conditions, the domain of \(\mathbf{r}(t)\) is \(t > 2\) and \(t \neq 4\), or in interval notation: \((2, 4) \cup (4, \infty)\).
**Range:**
To determine the range, analyze the output of each component based on the domain:
1. **\(\ln(t-2)\):**
- As \(t > 2\), \(\ln(t-2)\) can take any real number value from \(-\infty\) to \(\infty\).
2. **\(\frac{1}{t-4}\):**
- As \(t\) approaches 4 from either side, \(\frac{1}{t-4}\) approaches \(\pm \infty\).
- Thus, it can take any real number except 0.
3. **\(\sqrt{t}\):**
- Since \(t > 2\), \(\sqrt{t} > \sqrt{2}\).
Overall, the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8327d6e3-5c77-497b-a90c-d5ce68ac92e0%2Fe7fd6f9b-ff57-405d-9821-6f197f9fc7ac%2F0anvg2l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Determine the domain and range of the vector-valued function:
\[
\mathbf{r}(t) = \left\langle \ln(t-2), \frac{1}{t-4}, \sqrt{t} \right\rangle
\]
**Explanation:**
To find the domain of the vector-valued function \(\mathbf{r}(t)\), we need to determine the values of \(t\) for which each component of the function is defined.
1. **First Component \(\ln(t-2)\):**
- The logarithm function \(\ln(x)\) is defined for \(x > 0\).
- Thus, \(t-2 > 0 \Rightarrow t > 2\).
2. **Second Component \(\frac{1}{t-4}\):**
- The expression \(\frac{1}{x}\) is defined for \(x \neq 0\).
- Thus, \(t - 4 \neq 0 \Rightarrow t \neq 4\).
3. **Third Component \(\sqrt{t}\):**
- The square root function \(\sqrt{x}\) is defined for \(x \geq 0\).
- Thus, \(t \geq 0\).
Combining these conditions, the domain of \(\mathbf{r}(t)\) is \(t > 2\) and \(t \neq 4\), or in interval notation: \((2, 4) \cup (4, \infty)\).
**Range:**
To determine the range, analyze the output of each component based on the domain:
1. **\(\ln(t-2)\):**
- As \(t > 2\), \(\ln(t-2)\) can take any real number value from \(-\infty\) to \(\infty\).
2. **\(\frac{1}{t-4}\):**
- As \(t\) approaches 4 from either side, \(\frac{1}{t-4}\) approaches \(\pm \infty\).
- Thus, it can take any real number except 0.
3. **\(\sqrt{t}\):**
- Since \(t > 2\), \(\sqrt{t} > \sqrt{2}\).
Overall, the
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