Condition of existence of a function

Under the domain of real numbers (R) the function is defined if x>=0 or x<-1.

[f(x)]^2=x/(x+1)

at x=-1 you have division by zero. at x=0 f(x)=0

Then plot the function as (x,f(x)) and show that as x approaches - infinity f(x) approaches 1 and as x goes to -1 f(x) goes to infinity. At x=0 f(x) is zero and as x approaches + infinity f(x) goes to 1.

Between -1 and 0 the function isn't defined (in real numbers).

Rap

Assuming a forced choice alternative why not simply reject 1. on the grounds of tautology
If x>=0 then x+1 MUST BE > 0.

Alternatively complex numbers can be avoided in 2. by pointing out that if x=-1 then x+1=0 which would give a zero denominator.

Dear Fesco,

1. is based on condition of existence of sqrt(x) and sqrt(x + 1). If 1. is used the function is defined between for x>= 0.

2. is based on condition of existence of sqrt(x/(x+1)). If 1. is used the function is defined for x < -1 and x >= 0.

My sun was proposing 2. His teacher is proposing 1. So the problem is: can one insist that f(x) = sqrt(x) / sqrt(x + 1) differs from g(x) = sqrt(x/(x + 1)) (especially for x < -1)?