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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 nonzero. 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 x1> 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:$\displaystyle \{ x1\le x< 0\}\cup \{ x x> 1\}$. Last edited by Country Boy; October 25th, 2016 at 07:17 AM. 
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? 
October 25th, 2016, 11:34 PM  #4  
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry  Quote:  

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area, definition, find, function 
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