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Does the following represents function from R->R ?

 

f(x)=√x

In ths solution ,it is given:- f(x) is not defined for x < O. 

But do we consider it by default as +/-√x or is it only +√x?

I have asked a similar ques:-http://gateoverflow.in/132553/discrete-maths-group-theory

asked in Mathematical Logic by Veteran (15k points) 13 109 316 | 36 views
I think by default it will be $\sqrt{}$x unless implicitly specified as -$\sqrt{}$x

Also both $\sqrt{}$x and -$\sqrt{}$x will be undefined for x < 0.

2 Answers

+1 vote
if we consider f(x) = +or - sqrt(x) then it canot be a function because x can not take two different value at same time according to definition of function
answered by Active (2.2k points) 4 10
But if they have not mentioned any sign before the root,then can we assume default it will be only positive and hence a function
0 votes
$R \to R$ is mentioned in question meaning the domain and range of $f$ are the set of all real numbers. Now, $f(-2) = \sqrt{-2}$ is not a real number and thus $f$ is no longer a valid function.
answered by Veteran (319k points) 580 1448 2962
Sir,If it says Domain and range as R+->R. ,then can i say it is a function?

For  F(4),then my output will be only 2 or will it be +2 /-2 ?
@rahul It should be +2 only. As we doesn't have -2 in our domain.
But +2 and -2 are in range.Right?
Yes, just domain alone being positive is not enough for being a function - it then violates the rule that every point can have only one mapping.
Sir.but they have not mentioned +- explicitly with the output of the function.So shall i consider it will be +2 and -2 or shall we consider only +2?

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