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Get Full Access to Calculus: Early Transcendental Functions - 6 Edition - Chapter 9.9 - Problem 3
Get Full Access to Calculus: Early Transcendental Functions - 6 Edition - Chapter 9.9 - Problem 3

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# ?In Exercises 1-4, find a geometric power series for the function, centered at 0, (a) by the technique shown in Examples 1 and 2, and (b) by long divis

ISBN: 9781285774770 141

## Solution for problem 3 Chapter 9.9

Calculus: Early Transcendental Functions | 6th Edition

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Problem 3

In Exercises 1-4, find a geometric power series for the function, centered at 0, (a) by the technique shown in Examples 1 and 2, and (b) by long division.

$$f(x)=\frac{4}{3+x}$$

Text Transcription:

f(x) = 4 / 3 + x

Step-by-Step Solution:

Step 1 of 5) So 1>x is the derivative of ln x on the domain x 7 0, and the derivative of ln (-x) on the domain x 6 0.

Step 2 of 2

##### ISBN: 9781285774770

The full step-by-step solution to problem: 3 from chapter: 9.9 was answered by , our top Calculus solution expert on 11/14/17, 10:53PM. Since the solution to 3 from 9.9 chapter was answered, more than 255 students have viewed the full step-by-step answer. This full solution covers the following key subjects: . This expansive textbook survival guide covers 134 chapters, and 10738 solutions. The answer to “?In Exercises 1-4, find a geometric power series for the function, centered at 0, (a) by the technique shown in Examples 1 and 2, and (b) by long division.$$f(x)=\frac{4}{3+x}$$Text Transcription:f(x) = 4 / 3 + x” is broken down into a number of easy to follow steps, and 36 words. This textbook survival guide was created for the textbook: Calculus: Early Transcendental Functions, edition: 6. Calculus: Early Transcendental Functions was written by and is associated to the ISBN: 9781285774770.

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