My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum

LinkBack Thread Tools Display Modes
October 25th, 2016, 06:19 AM   #1
Joined: Oct 2016
From: Indonesia

Posts: 1
Thanks: 0

Post find the definition area of the function

noreason is offline  
October 25th, 2016, 07:12 AM   #2
Math Team
Joined: Jan 2015
From: Alabama

Posts: 3,264
Thanks: 902

"Definition area"? Do you mean the "natural domain" of the function- the region in which the formula is defined? In order that a number have a square root that number must be non-zero. Here, we must have $\displaystyle x- \frac{1}{x}\ge 0$.

If x is positive, multiplying both sides by x we have $\displaystyle x^2- 1= (x- 1)(x+ 1)\ge 0$. In order for that to be true, both x- 1 and x+ 1 must have the same sign: either x-1> 0 and x+ 1> 0 or x- 1< 0 and x+ 1< 0. The first pair of inequalities is true for x> 1 and the second for x< -1. Since "x is positive", we must have x> 1.

If x is negative, multiplying both sides by x we have $\displaystyle x^2- 1= (x- 1)(x+ 1)\ge 0$. In order for that to be true, x- 1 and x+ 1 must have opposite signs: x+ 1> 0 and x- 1< 0 or x- 1<0 and x- 1> 0. The first pair is true for $\displaystyle -1\le 0 \le 1$. The second pair are never both true. Since "x is negative" we must have $\displaystyle -1\le x\le 0$.

Of course, we also cannot divide by 0 so x= 0 is not in the domain. The domain is the union of the two separate sets:$\displaystyle \{ x|-1\le x< 0\}\cup \{ x| x> 1\}$.

Last edited by Country Boy; October 25th, 2016 at 07:17 AM.
Country Boy is offline  
October 25th, 2016, 07:17 AM   #3
Senior Member
Joined: May 2016
From: USA

Posts: 1,310
Thanks: 551

It is actually impossible to answer this question because it is ambiguous. If you are talking about real functions, there is one answer, and if you are talking about complex functions, there is a different answer.

Let's suppose we are talking about real functions of real variables, and let's define

$g(x) = x - \dfrac{1}{x}\ and\ f(x) = \sqrt{g(x)}\ such\ that\ x,\ g(x),\ f(x) \in \mathbb R.$

For what values of g(x) is f(x) a real number?

Consequently, for what values of x is f(x) a real number?
JeffM1 is offline  
October 25th, 2016, 11:34 PM   #4
Senior Member
Monox D. I-Fly's Avatar
Joined: Nov 2010
From: Indonesia

Posts: 2,001
Thanks: 132

Math Focus: Trigonometry
Originally Posted by noreason View Post
Hi fellow Indonesian! To be honest, we don't know what you are asking for. Maybe you can post the question in our native language so I can tell the others what do you intend to ask?
Monox D. I-Fly is offline  

  My Math Forum > College Math Forum > Calculus

area, definition, find, function

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Is It Possible To Write This As A Function With One Definition? EvanJ Pre-Calculus 3 October 6th, 2015 01:28 PM
How to find derivative by definition? rain Calculus 4 November 2nd, 2013 01:45 PM
Problem on function definition yugimutoshung Algebra 2 April 26th, 2013 09:21 PM
how to find area between function ? safyras Calculus 2 December 5th, 2010 08:25 AM
Area under a curve, limit definition Andrey Calculus 1 February 3rd, 2008 09:45 AM

Copyright © 2019 My Math Forum. All rights reserved.