Show that the function f defined by
$f(x):= \dfrac{x}{\sqrt{x^2+1}}$ $x \in R$, is a bijection of R onto {y: -1

Question

Show that the function f defined by 
$f(x):= \dfrac{x}{\sqrt{x^2+1}}$ $x \in R$, is a bijection of R onto {y: -1<y<1}
Please, help.

loser66 · Answer

to prove f(x) is a bijection, we need prove if $f(x_1)\neq f(x_2)$ , then $x_1 \neq x_2$
Assume $f(x_1)= f(x_2)$ then 
$$\dfrac{x_1}{\sqrt{x_1^2+1}}=\dfrac{x_2}{\sqrt{x_2^2+1}}$$
After solving, I get $x_1^2 = x_2^2$ which gives me $\pm x_1=\pm x_2$ . Then I am stuck.

loser66 · Answer

@ganeshie8 @cwrw238

kainui · Answer

Why are you assuming f(x1)=f(x2) if you need to prove that they aren't equal? By assuming they're not equal then you will get the result that x1 does not equal x2 right? Which is what you've set out to prove.

loser66 · Answer

To prove it is a bijection, we need to prove it is injective and then surjective,
I know how to do surjective part. 
for injective, by definition, if x1 $\neq x_2~~ then~~ f(x_1)\neq f(x_2$ . I use contrapositive method by assume f(x1) = f(x2) and let to x1 =x2

loser66 · Answer

@phi

phi · Answer

I would simplify your result to 
$$ x_1= \pm x_2 $$
If $x_1 = x_2$  then you have the case where f(x1) = f(x2) when x1= x2
if you have x1= -x2   then  f(x1) ≠ f(x2)  when x1 ≠ x2

loser66 · Answer

Thank you so much. I got it