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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Hi, I was wondering where the 5 in the denominator came from. This is supposed to be the answer so I am trying to understand the work. Thanks!
![### Continuity of Piecewise Functions
Consider the function \( g(t) \) defined piecewise as follows:
\[
g(t) =
\begin{cases}
-t & \text{if } t < 0 \\
\sin^{-1}(t) + 3a & \text{if } 0 \leq t < 1 \\
5t + \frac{b}{t} & \text{if } t \geq 1
\end{cases}
\]
where \( a \) and \( b \) are constants.
#### Problem Statement
**(a)** Find \( a \) and \( b \) such that \( g(t) \) is continuous on \( (-\infty, \infty) \).
#### Solution
To ensure \( g(t) \) is continuous on \( (-\infty, \infty) \), \( g(t) \) must be continuous at the points where the definition of \( g(t) \) changes, i.e., at \( t = 0 \) and \( t = 1 \).
**1. Continuity at \( t = 0 \):**
We need:
\[ \lim_{t \to 0} g(t) = g(0) \]
First, calculate:
\[ \lim_{t \to 0^-} g(t) = \lim_{t \to 0^-} (-t) = 0 \]
\[ g(0) = \sin^{-1}(0) + 3a = 0 + 3a = 3a \]
For continuity at \( t = 0 \):
\[ \lim_{t \to 0} g(t) = 3a \]
Since:
\[ \lim_{t \to 0} (-t) = 0 \]
We get:
\[ 3a = 0 \]
Thus, \( a = 0 \).
**2. Continuity at \( t = 1 \):**
We need:
\[ \lim_{t \to 1} g(t) = g(1) \]
First, calculate:
\[ \lim_{t \to 1^-} \sin^{-1}(t) + 3a \]
Since \( a = 0 \):
\[ \lim_{t \to 1^-} \sin^{-1}(t) = \sin^{-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Febb676ba-83b9-4d8c-bd2c-0a9e39a90a71%2F046c9506-516c-41b0-9623-36063a098bc8%2Fp1djrgp_processed.png&w=3840&q=75)
Transcribed Image Text:### Continuity of Piecewise Functions
Consider the function \( g(t) \) defined piecewise as follows:
\[
g(t) =
\begin{cases}
-t & \text{if } t < 0 \\
\sin^{-1}(t) + 3a & \text{if } 0 \leq t < 1 \\
5t + \frac{b}{t} & \text{if } t \geq 1
\end{cases}
\]
where \( a \) and \( b \) are constants.
#### Problem Statement
**(a)** Find \( a \) and \( b \) such that \( g(t) \) is continuous on \( (-\infty, \infty) \).
#### Solution
To ensure \( g(t) \) is continuous on \( (-\infty, \infty) \), \( g(t) \) must be continuous at the points where the definition of \( g(t) \) changes, i.e., at \( t = 0 \) and \( t = 1 \).
**1. Continuity at \( t = 0 \):**
We need:
\[ \lim_{t \to 0} g(t) = g(0) \]
First, calculate:
\[ \lim_{t \to 0^-} g(t) = \lim_{t \to 0^-} (-t) = 0 \]
\[ g(0) = \sin^{-1}(0) + 3a = 0 + 3a = 3a \]
For continuity at \( t = 0 \):
\[ \lim_{t \to 0} g(t) = 3a \]
Since:
\[ \lim_{t \to 0} (-t) = 0 \]
We get:
\[ 3a = 0 \]
Thus, \( a = 0 \).
**2. Continuity at \( t = 1 \):**
We need:
\[ \lim_{t \to 1} g(t) = g(1) \]
First, calculate:
\[ \lim_{t \to 1^-} \sin^{-1}(t) + 3a \]
Since \( a = 0 \):
\[ \lim_{t \to 1^-} \sin^{-1}(t) = \sin^{-
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