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?In Exercises 1–6, match the equation with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] \(\

Calculus: Early Transcendental Functions | 6th Edition | ISBN: 9781285774770 | Authors: Ron Larson ISBN: 9781285774770 141

Solution for problem 4 Chapter 10.1

Calculus: Early Transcendental Functions | 6th Edition

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Calculus: Early Transcendental Functions | 6th Edition | ISBN: 9781285774770 | Authors: Ron Larson

Calculus: Early Transcendental Functions | 6th Edition

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Problem 4

In Exercises 1–6, match the equation with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).]

\(\frac{(x-2)^{2}}{16}+\frac{(y+1)^{2}}{4}=1\)

Text Transcription:

(x-2)^\2 / 16 + (y + 1)^2 / 4 = 1

Step-by-Step Solution:

Step 1 of 5) Proof Part (1). If ƒ(c) 6 0, then ƒ(x) 6 0 on some open interval I containing the point c, since ƒ is continuous. Therefore, ƒ is decreasing on I. Since ƒ(c) = 0, the sign of ƒ changes from positive to negative at c so ƒ has a local maximum at c by the First Derivative Test.

Step 2 of 2

Chapter 10.1, Problem 4 is Solved
Textbook: Calculus: Early Transcendental Functions
Edition: 6
Author: Ron Larson
ISBN: 9781285774770

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?In Exercises 1–6, match the equation with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] \(\