College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
Problem 84E
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![### Understanding Continuity and Discontinuity of Functions
#### Discontinuity in Function \( h \)
Consider the function \( h(x) \):
\[ h(x) =
\begin{cases}
\frac{1}{x + 2} & \text{if } x \ne -2 \\
1 & \text{if } x = -2
\end{cases} \]
**Discontinuity Explanation:**
To determine why the function \( h \) is discontinuous at \( a = -2 \):
1. **Limit from the Left and Right:**
- As \( x \) approaches \(-2\) from either direction (left or right), \( \frac{1}{x + 2} \) becomes undefined. However, for values very close to but not equal to \(-2\), the function yields values approaching \( \pm \infty \).
2. **Value at \( x = -2 \):**
- When \( x \) reaches \(-2\), \( h(x) \) is defined as \( 1 \), which is explicitly given to avoid the undefined nature of \( \frac{1}{x + 2} \).
Since \( \lim_{{x \to -2}} h(x) \) does not equal \( h(-2) \), the function \( h \) is discontinuous at \( x = -2 \).
#### Continuity in Function \( f \)
Consider the function \( f(x) \):
\[ f(x) = \frac{3v - 1}{v^2 + 2v - 15} \]
**Continuity and Domain:**
1. **Finding the Domain:**
- The denominator \( v^2 + 2v - 15 \) should not be zero to avoid undefined values.
- Solving \( v^2 + 2v - 15 = 0 \):
- \( v^2 + 5v - 3v - 15 = 0 \)
- \((v + 5)(v - 3) = 0\)
Thus, the denominator is zero when \( v = -5 \) or \( v = 3 \). Hence, the domain of \( f \) is all real numbers except \( v = -5 \) and \( v = 3 \).
2. **Checking for Continu](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc4971f23-5570-4fed-8cd6-5e1a07d18a42%2Fb2b6bc71-88a6-4019-8fd1-0376d6421c2d%2F0n5w066_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Continuity and Discontinuity of Functions
#### Discontinuity in Function \( h \)
Consider the function \( h(x) \):
\[ h(x) =
\begin{cases}
\frac{1}{x + 2} & \text{if } x \ne -2 \\
1 & \text{if } x = -2
\end{cases} \]
**Discontinuity Explanation:**
To determine why the function \( h \) is discontinuous at \( a = -2 \):
1. **Limit from the Left and Right:**
- As \( x \) approaches \(-2\) from either direction (left or right), \( \frac{1}{x + 2} \) becomes undefined. However, for values very close to but not equal to \(-2\), the function yields values approaching \( \pm \infty \).
2. **Value at \( x = -2 \):**
- When \( x \) reaches \(-2\), \( h(x) \) is defined as \( 1 \), which is explicitly given to avoid the undefined nature of \( \frac{1}{x + 2} \).
Since \( \lim_{{x \to -2}} h(x) \) does not equal \( h(-2) \), the function \( h \) is discontinuous at \( x = -2 \).
#### Continuity in Function \( f \)
Consider the function \( f(x) \):
\[ f(x) = \frac{3v - 1}{v^2 + 2v - 15} \]
**Continuity and Domain:**
1. **Finding the Domain:**
- The denominator \( v^2 + 2v - 15 \) should not be zero to avoid undefined values.
- Solving \( v^2 + 2v - 15 = 0 \):
- \( v^2 + 5v - 3v - 15 = 0 \)
- \((v + 5)(v - 3) = 0\)
Thus, the denominator is zero when \( v = -5 \) or \( v = 3 \). Hence, the domain of \( f \) is all real numbers except \( v = -5 \) and \( v = 3 \).
2. **Checking for Continu
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![### Continuity of the Function \( f(v) \)
**Problem Statement:**
Explain why the function \( f \) is continuous at every number in its domain. State the domain.
\[ f(v) = \frac{3v - 1}{v^2 + 2v - 15} \]
**Solution Explanation:**
To determine the continuity of the function \( f \) at every number in its domain, first, we need to identify the domain of the function. The function \( f \) is given as a rational function. Rational functions are continuous everywhere they are defined. The function is defined for all real numbers \( v \) except where the denominator is zero.
1. **Find the Domain:**
To find where the denominator is zero, solve the equation \( v^2 + 2v - 15 = 0 \):
\[
v^2 + 2v - 15 = 0
\]
Factorize the quadratic expression:
\[
(v + 5)(v - 3) = 0
\]
Set each factor equal to zero:
\[
v + 5 = 0 \quad \Rightarrow \quad v = -5
\]
\[
v - 3 = 0 \quad \Rightarrow \quad v = 3
\]
Therefore, the denominator is zero at \( v = -5 \) and \( v = 3 \).
2. **State the Domain:**
Since the function \( f(v) \) is undefined where the denominator is zero, the domain of \( f \) is all real numbers except \( v = -5 \) and \( v = 3 \):
\[
\text{Domain: } v \in \mathbb{R} \setminus \{-5, 3\}
\]
3. **Continuity:**
The function \( f(v) \) is continuous at every point in its domain. This is because for rational functions, continuity is determined by the denominator. As long as the denominator is not zero, the function is continuous. So, for all \( v \) in the domain \( \mathbb{R} \setminus \{-5, 3\} \), the function is continuous.
In conclusion, the function \( f(v) \) is continuous at every](https://content.bartleby.com/qna-images/question/c4971f23-5570-4fed-8cd6-5e1a07d18a42/105fd064-a2b0-4b3c-acf0-bf18aa294fa7/e7bi1wo_processed.png)
Transcribed Image Text:### Continuity of the Function \( f(v) \)
**Problem Statement:**
Explain why the function \( f \) is continuous at every number in its domain. State the domain.
\[ f(v) = \frac{3v - 1}{v^2 + 2v - 15} \]
**Solution Explanation:**
To determine the continuity of the function \( f \) at every number in its domain, first, we need to identify the domain of the function. The function \( f \) is given as a rational function. Rational functions are continuous everywhere they are defined. The function is defined for all real numbers \( v \) except where the denominator is zero.
1. **Find the Domain:**
To find where the denominator is zero, solve the equation \( v^2 + 2v - 15 = 0 \):
\[
v^2 + 2v - 15 = 0
\]
Factorize the quadratic expression:
\[
(v + 5)(v - 3) = 0
\]
Set each factor equal to zero:
\[
v + 5 = 0 \quad \Rightarrow \quad v = -5
\]
\[
v - 3 = 0 \quad \Rightarrow \quad v = 3
\]
Therefore, the denominator is zero at \( v = -5 \) and \( v = 3 \).
2. **State the Domain:**
Since the function \( f(v) \) is undefined where the denominator is zero, the domain of \( f \) is all real numbers except \( v = -5 \) and \( v = 3 \):
\[
\text{Domain: } v \in \mathbb{R} \setminus \{-5, 3\}
\]
3. **Continuity:**
The function \( f(v) \) is continuous at every point in its domain. This is because for rational functions, continuity is determined by the denominator. As long as the denominator is not zero, the function is continuous. So, for all \( v \) in the domain \( \mathbb{R} \setminus \{-5, 3\} \), the function is continuous.
In conclusion, the function \( f(v) \) is continuous at every
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