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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 1.1 - Problem 51e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 1.1 - Problem 51e

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# x2/3+y 2/3= 1

ISBN: 9780321570567 2

## Solution for problem 51E Chapter 1.1

Calculus: Early Transcendentals | 1st Edition

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Problem 51E

x2/3+y 2/3= 1

Step-by-Step Solution:

Step by step solution Step 1 of 2 Given relation x2/+y 2/3= 1. Let’s rearrange this relation: x2/+y 2/3= 1 y 2/3= 1x 2/3 y = (1x )2/33/2 2/3 y = (1x ) Therefore we have the function : 3 y = f(x) = (1x )2/3 This function is square root function and for any square root function the quantity under the root must be nonnegative. This means that: (1x ) 0 1x 2/3 0 1 x 1 (the function (x) is only defined for all real numbers between -1 and 1) Let s find the value (-x): 3 f(x) = (1(x) )/3 2/33 f(x) = (1(x) ) 2/3 3 2 3 2 3 2 If we write (x) as (x) then we can noticed that (x) = (+x) . Therefore we have: f(x) = (1(x) ) =3 (1(+x) )/33 f(x) = f(x) Based on the previous we an conclude that the funct (x) is even and it is symmetric with respect to th -axis. If the function is symmetric with respect to the y-axis then its values for and x are equal. Let’s look the the graph of f(x)..

Step 2 of 2

##### ISBN: 9780321570567

The full step-by-step solution to problem: 51E from chapter: 1.1 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. Since the solution to 51E from 1.1 chapter was answered, more than 390 students have viewed the full step-by-step answer. The answer to “x2/3+y 2/3= 1” is broken down into a number of easy to follow steps, and 3 words. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. This full solution covers the following key subjects: . This expansive textbook survival guide covers 85 chapters, and 5218 solutions.

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