Very simple applications' issue

[Andrax]

Homework Statement

so i want to prove that an application is bijective
y [itex]\geq[/itex] 1we're looking for an x[itex]\geq[/itex]-2 : y= f(x)
anyway at the end
i have lx+2l=[itex]\sqrt{}y+1[/itex]
the teached said x = [itex]\sqrt{}y+1[/itex]-2 without studying the other case
Ps: the +1 is includedi nthe square root

Homework Equations

The Attempt at a Solution

the other case lx+2l=[itex]\sqrt{}y+1[/itex]
x+2=[itex]\sqrt{}y+1[/itex] or x+2=-[itex]\sqrt{}y+1[/itex]

the second x is going to be <2 so we can exclude it the first x is going to be >2 we're oging to use it , but why our teacher didn't study the second case in which x+2 = -[itex]\sqrt{}y+1[/itex] ?

 

[Mark44]

Homework Statement

so i want to prove that an application is bijective
y [itex]\in[/itex] [itex]\left[1[/itex](y >=1)+[itex]\infty[/itex][itex]\left[[/itex] we're looking for an x from [itex]\left[-2[/itex]+[itex]\infty[/itex][itex]\left[[/itex](x>=-2) : y= f(x)
anyway at the end
i have lx+2l=[itex]\sqrt{}y+1[/itex]
the teached said x = [itex]\sqrt{}y+1[/itex]-2 without studying the other case
Ps: the +1 is includedi nthe square root

Homework Equations

Your LaTeX is broken, so I can't understand what you have attempted to write.

The Attempt at a Solution

the other case lx+2l=[itex]\sqrt{}y+1[/itex]
x+2=[itex]\sqrt{}y+1[/itex] or x+2=-[itex]\sqrt{}y+1[/itex]

the second x is going to be <2 so we can exclude it the first x is going to be >2 we're oging to use it , but why our teacher didn't study the second case in which x+2 = -[itex]\sqrt{}y+1[/itex] ?