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9. Find the value of a so that the function g(x) is continuous at x = 0 where g(x) = { = 1 2xe3x 1 x(x + 2) if x #0 if x = 0

9. Find the value of a so that the function g(x) is continuous at x = 0 where g(x) = { = 1 2xe3x 1 x(x + 2) if x #0 if x = 0

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 9**: Find the value of \( a \) so that the function \( g(x) \) is continuous at \( x = 0 \) where \[ g(x) = \begin{cases} \frac{1}{2x e^{3x}} - \frac{1}{x(x+2)} & \text{if } x \neq 0 \\ a & \text{if } x = 0 \end{cases} \] In order to ensure that \( g(x) \) is continuous at \( x = 0 \), the limit of the function as \( x \) approaches 0 from both sides must equal \( g(0) \). Therefore, calculate the limit of \( \frac{1}{2x e^{3x}} - \frac{1}{x(x+2)} \) as \( x \to 0 \) and set it equal to \( a \).
Transcribed Image Text:**Problem 9**: Find the value of \( a \) so that the function \( g(x) \) is continuous at \( x = 0 \) where \[ g(x) = \begin{cases} \frac{1}{2x e^{3x}} - \frac{1}{x(x+2)} & \text{if } x \neq 0 \\ a & \text{if } x = 0 \end{cases} \] In order to ensure that \( g(x) \) is continuous at \( x = 0 \), the limit of the function as \( x \) approaches 0 from both sides must equal \( g(0) \). Therefore, calculate the limit of \( \frac{1}{2x e^{3x}} - \frac{1}{x(x+2)} \) as \( x \to 0 \) and set it equal to \( a \).
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