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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

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.

+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 ago by Active (1.3k points)
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
$R \to R$ 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 ago by Veteran (286k points)
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?