let f be the function x²-C for x< 5 flx)=3x+5c for X Z 5

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...
Question
**Continuity of Piecewise Functions**

In calculus, it's essential to understand how to make piecewise functions continuous at the junction points. 

Consider the following piecewise function \( f \):

\[ f(x) = \begin{cases} 
x^2 - c & \text{for } x < 5 \\
3x + 5c & \text{for } x \geq 5 
\end{cases} \]

We are tasked with finding the value of \( c \) that ensures \( f \) is continuous across its entire domain.

---

To check for continuity at \( x = 5 \):
1. Ensure the left-hand limit is equal to the right-hand limit at \( x = 5 \).
2. Ensure the function value at \( x = 5 \) matches these limits.

\[ \lim_{{x \to 5^-}} f(x) = \lim_{{x \to 5^+}} f(x) = f(5) \]

Let's compute these limits and function values:

### Left-hand limit as \( x \) approaches 5:
\[ \lim_{{x \to 5^-}} f(x) = \lim_{{x \to 5^-}} (x^2 - c) = 5^2 - c = 25 - c \]

### Right-hand limit as \( x \) approaches 5:
\[ \lim_{{x \to 5^+}} f(x) = \lim_{{x \to 5^+}} (3x + 5c) = 3(5) + 5c = 15 + 5c \]

### Ensure the function is continuous:
Set the left-hand limit equal to the right-hand limit:
\[ 25 - c = 15 + 5c \]

### Solve for \( c \):
\[ 25 - 15 = 5c + c \]
\[ 10 = 6c \]
\[ c = \frac{10}{6} = \frac{5}{3} \]

Therefore, the value of \( c \) that makes the function continuous is \( \frac{5}{3} \). 

--- 

This approach illustrates how you can analytically find the necessary conditions to maintain the continuity of piecewise functions.
Transcribed Image Text:**Continuity of Piecewise Functions** In calculus, it's essential to understand how to make piecewise functions continuous at the junction points. Consider the following piecewise function \( f \): \[ f(x) = \begin{cases} x^2 - c & \text{for } x < 5 \\ 3x + 5c & \text{for } x \geq 5 \end{cases} \] We are tasked with finding the value of \( c \) that ensures \( f \) is continuous across its entire domain. --- To check for continuity at \( x = 5 \): 1. Ensure the left-hand limit is equal to the right-hand limit at \( x = 5 \). 2. Ensure the function value at \( x = 5 \) matches these limits. \[ \lim_{{x \to 5^-}} f(x) = \lim_{{x \to 5^+}} f(x) = f(5) \] Let's compute these limits and function values: ### Left-hand limit as \( x \) approaches 5: \[ \lim_{{x \to 5^-}} f(x) = \lim_{{x \to 5^-}} (x^2 - c) = 5^2 - c = 25 - c \] ### Right-hand limit as \( x \) approaches 5: \[ \lim_{{x \to 5^+}} f(x) = \lim_{{x \to 5^+}} (3x + 5c) = 3(5) + 5c = 15 + 5c \] ### Ensure the function is continuous: Set the left-hand limit equal to the right-hand limit: \[ 25 - c = 15 + 5c \] ### Solve for \( c \): \[ 25 - 15 = 5c + c \] \[ 10 = 6c \] \[ c = \frac{10}{6} = \frac{5}{3} \] Therefore, the value of \( c \) that makes the function continuous is \( \frac{5}{3} \). --- This approach illustrates how you can analytically find the necessary conditions to maintain the continuity of piecewise functions.
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