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**g(x) = [tex]\frac{2\left|x\right|}{\sqrt{x^2 +1}}[/tex]**

Without sketching any graphs, show that g is NOT one to one.

Without sketching any graphs, show that g is NOT one to one.

**3. The Attempt at a Solution**

Well, I know that in this situation it is required to show that if [tex]f(x_{1}) = f(x_{2})[/tex], then [tex]x_{1} = x_{2}[/tex]

[tex]f(x_{1}) = f(x_{2})[/tex] => [tex]\frac{2\left|x_{1}\right|}{\sqrt{x_{1}^2 +1}}[/tex] = [tex]\frac{2\left|x_{2}\right|}{\sqrt{x_{2}^2 +1}}[/tex]

This is interesting! but I'm not quite sure what I should do next...

[tex](2\left|x_{1}\right|) (\sqrt{x_{2}^2 +1})= (2\left|x_{2}\right|) (\sqrt{x_{1}^2 +1})[/tex] ?

Does anybody know how to complete this proof?

Well, I know that in this situation it is required to show that if [tex]f(x_{1}) = f(x_{2})[/tex], then [tex]x_{1} = x_{2}[/tex]

[tex]f(x_{1}) = f(x_{2})[/tex] => [tex]\frac{2\left|x_{1}\right|}{\sqrt{x_{1}^2 +1}}[/tex] = [tex]\frac{2\left|x_{2}\right|}{\sqrt{x_{2}^2 +1}}[/tex]

This is interesting! but I'm not quite sure what I should do next...

[tex](2\left|x_{1}\right|) (\sqrt{x_{2}^2 +1})= (2\left|x_{2}\right|) (\sqrt{x_{1}^2 +1})[/tex] ?

Does anybody know how to complete this proof?

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