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
![### Piecewise Function and Continuity
**Problem Statement:**
The function is given piecewise as follows:
\[
f(x) =
\begin{cases}
e^x & \text{if } x < 0 \\
0 & \text{if } x = 0 \\
\tan(x) & \text{if } x > 0
\end{cases}
\]
**Question:**
Is the function continuous at \( x = 0 \)? Explain.
**Solution:**
To determine if the function is continuous at \( x = 0 \), we need to check the following three conditions of continuity at a point \( c \):
1. \( f(c) \) is defined.
2. The limit of \( f(x) \) as \( x \) approaches \( c \) exists.
3. The limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \).
Let's check each of these conditions for \( c = 0 \):
1. **Is \( f(0) \) defined?**
Yes, \( f(0) = 0 \).
2. **Does \( \lim_{x \to 0} f(x) \) exist?**
We need to check the left-hand limit and the right-hand limit:
- **Left-hand limit:**
\[
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^x = e^0 = 1
\]
- **Right-hand limit:**
\[
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \tan(x)
\]
The limit of \( \tan(x) \) as \( x \) approaches 0 from the right is 0.
Since the left-hand limit and the right-hand limit are not equal (1 versus 0), the limit of \( f(x) \) as \( x \) approaches 0 does not exist.
3. **Is \( \lim_{x \to 0} f(x) = f(0) \)?**
Since the limit does not exist, this condition is not satisfied.
Therefore, the function \( f(x) \) is **not continuous** at \( x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63c12736-f43f-4777-9d94-03ebf9ed40f6%2F289a5cd4-e1ae-4780-a303-db509ee75c9a%2Fd7yxv5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Piecewise Function and Continuity
**Problem Statement:**
The function is given piecewise as follows:
\[
f(x) =
\begin{cases}
e^x & \text{if } x < 0 \\
0 & \text{if } x = 0 \\
\tan(x) & \text{if } x > 0
\end{cases}
\]
**Question:**
Is the function continuous at \( x = 0 \)? Explain.
**Solution:**
To determine if the function is continuous at \( x = 0 \), we need to check the following three conditions of continuity at a point \( c \):
1. \( f(c) \) is defined.
2. The limit of \( f(x) \) as \( x \) approaches \( c \) exists.
3. The limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \).
Let's check each of these conditions for \( c = 0 \):
1. **Is \( f(0) \) defined?**
Yes, \( f(0) = 0 \).
2. **Does \( \lim_{x \to 0} f(x) \) exist?**
We need to check the left-hand limit and the right-hand limit:
- **Left-hand limit:**
\[
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^x = e^0 = 1
\]
- **Right-hand limit:**
\[
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \tan(x)
\]
The limit of \( \tan(x) \) as \( x \) approaches 0 from the right is 0.
Since the left-hand limit and the right-hand limit are not equal (1 versus 0), the limit of \( f(x) \) as \( x \) approaches 0 does not exist.
3. **Is \( \lim_{x \to 0} f(x) = f(0) \)?**
Since the limit does not exist, this condition is not satisfied.
Therefore, the function \( f(x) \) is **not continuous** at \( x
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