Very simple applications' issue

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Homework Statement


so i want to prove that an application is bijective
y \geq 1we're looking for an x\geq-2 : y= f(x)
anyway at the end
i have lx+2l=\sqrt{}y+1
the teached said x = \sqrt{}y+1-2 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=\sqrt{}y+1
x+2=\sqrt{}y+1 or x+2=-\sqrt{}y+1

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 = -\sqrt{}y+1 ?
 
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Andrax said:

Homework Statement


so i want to prove that an application is bijective
y \in \left[1(y >=1)+\infty\left[ we're looking for an x from \left[-2+\infty\left[(x>=-2) : y= f(x)
anyway at the end
i have lx+2l=\sqrt{}y+1
the teached said x = \sqrt{}y+1-2 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.
Andrax said:

The Attempt at a Solution


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

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 = -\sqrt{}y+1 ?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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