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(b) Let f(x) = [(3/6=+1) x<1 2x+3b, x21 continuous everywhere. . Find the constant X=1 b such the function fis 1-1=R-L= f(x)

(b) Let f(x) = [(3/6=+1) x<1 2x+3b, x21 continuous everywhere. . Find the constant X=1 b such the function fis 1-1=R-L= f(x)

Calculus: Early Transcendentals
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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### Problem Statement **(b)** Let \( f(x) = \begin{cases} 3^{\frac{3}{x+1}} & , \, x < 1 \\ 2x + 3b & , \, x \geq 1 \end{cases} \). Find the constant \( b \) such that the function \( f \) is continuous everywhere. ### Explanation: To ensure that the function \( f \) is continuous everywhere, the following condition must be met: \[ \lim_{{x \to c^-}} f(x) = \lim_{{x \to c^+}} f(x) = f(c) \] For the point where \( x = 1 \): \[ x = 1 \] This means that: \[ \lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^+}} f(x) = f(1) \] Hence, the values from both piecewise sections at \( x = 1 \) need to be equal. \( 1. 1 = \text{RL} = \text{f(x)} \)
Transcribed Image Text:### Problem Statement **(b)** Let \( f(x) = \begin{cases} 3^{\frac{3}{x+1}} & , \, x < 1 \\ 2x + 3b & , \, x \geq 1 \end{cases} \). Find the constant \( b \) such that the function \( f \) is continuous everywhere. ### Explanation: To ensure that the function \( f \) is continuous everywhere, the following condition must be met: \[ \lim_{{x \to c^-}} f(x) = \lim_{{x \to c^+}} f(x) = f(c) \] For the point where \( x = 1 \): \[ x = 1 \] This means that: \[ \lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^+}} f(x) = f(1) \] Hence, the values from both piecewise sections at \( x = 1 \) need to be equal. \( 1. 1 = \text{RL} = \text{f(x)} \)
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