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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![**Topic: Continuity of Functions**
**Exercise 17: Redefine the Function for Continuity**
To ensure that the function \( f \) is continuous at \( x = 0 \), we must redefine it appropriately. A function \( f \) is given as:
\[
f(x) =
\begin{cases}
\frac{9}{x} + \frac{-8x + 36}{x(x-4)} & \text{if } x \ne 0, 4 \\
4 & \text{if } x = 0, 4
\end{cases}
\]
**Objective:**
Redefine the function \( f \) such that it remains continuous at \( x = 0 \).
**Explanation:**
The function \( f(x) \) has two parts. For values of \( x \) that are not 0 or 4, it takes the form \( \frac{9}{x} + \frac{-8x + 36}{x(x-4)} \). At \( x = 0 \) and \( x = 4 \), the function is given the value of 4.
To make \( f(x) \) continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 matches the value of the function at \( x = 0 \).
Hence, follow these steps for a solution:
1. Evaluate \( \lim_{{x \to 0}} f(x) \).
2. Ensure this limit equals \( f(0) \).
By solving this, we can redefine the function appropriately to maintain its continuity at \( x = 0 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd45308af-f507-4349-a094-dce43ad17ed2%2Fd6645d65-d8c7-4cc7-921b-a9c3b2612f3c%2Fesroy7b.png&w=3840&q=75)
Transcribed Image Text:**Topic: Continuity of Functions**
**Exercise 17: Redefine the Function for Continuity**
To ensure that the function \( f \) is continuous at \( x = 0 \), we must redefine it appropriately. A function \( f \) is given as:
\[
f(x) =
\begin{cases}
\frac{9}{x} + \frac{-8x + 36}{x(x-4)} & \text{if } x \ne 0, 4 \\
4 & \text{if } x = 0, 4
\end{cases}
\]
**Objective:**
Redefine the function \( f \) such that it remains continuous at \( x = 0 \).
**Explanation:**
The function \( f(x) \) has two parts. For values of \( x \) that are not 0 or 4, it takes the form \( \frac{9}{x} + \frac{-8x + 36}{x(x-4)} \). At \( x = 0 \) and \( x = 4 \), the function is given the value of 4.
To make \( f(x) \) continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 matches the value of the function at \( x = 0 \).
Hence, follow these steps for a solution:
1. Evaluate \( \lim_{{x \to 0}} f(x) \).
2. Ensure this limit equals \( f(0) \).
By solving this, we can redefine the function appropriately to maintain its continuity at \( x = 0 \).
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