1 votes 1 votes 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:-https://gateoverflow.in/132553/discrete-maths-group-theory Set Theory & Algebra discrete-mathematics + – rahul sharma 5 asked Jun 10, 2017 recategorized Mar 21, 2018 by srestha rahul sharma 5 610 views answer comment Share Follow See 1 comment See all 1 1 comment reply just_bhavana commented Jun 18, 2017 reply Follow Share 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. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes 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 Niraj Singh 2 answered Jun 19, 2017 Niraj Singh 2 comment Share Follow See 1 comment See all 1 1 comment reply rahul sharma 5 commented Jun 19, 2017 reply Follow Share 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 0 votes Please log in or register to add a comment.
1 votes 1 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. Arjun answered Jun 19, 2017 Arjun comment Share Follow See all 5 Comments See all 5 5 Comments reply rahul sharma 5 commented Jun 19, 2017 reply Follow Share 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 ? 0 votes 0 votes Hemant Parihar commented Jun 20, 2017 reply Follow Share @rahul It should be +2 only. As we doesn't have -2 in our domain. 0 votes 0 votes rahul sharma 5 commented Jun 20, 2017 reply Follow Share But +2 and -2 are in range.Right? 0 votes 0 votes Arjun commented Jun 20, 2017 reply Follow Share 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. 0 votes 0 votes rahul sharma 5 commented Jun 22, 2017 reply Follow Share 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? 0 votes 0 votes Please log in or register to add a comment.