Prove $f(x)=\frac{x}{\sqrt{x^2+1}}$ to be injective

Question

The function $f$ from the real set to $\mathopen]-1,1\mathclose[$ is defined as $$ f(x)=\frac{x}{\sqrt{x^2+1}} $$ I have to prove it injective.

I supposed there are two reals $a$ and $b$ such that $f(a)=f(b)$, but came to the result that $a=\pm b$.

What's wrong?

Keep going. show that if $a=-b$, then $f(a)$ does not equal $f(b)$ (unless $a=0$) — mathworker21, Sep 16 '17 at 09:31
Ah thanks. I saw that a=b=0 but didn't pay attention they'd be equal if so. — user480659, Sep 16 '17 at 09:39

score 2 · Accepted Answer · answered Sep 16 '17 at 09:33

2

If $a=-b$, $f(a) > 0$ and $f(b)<0$ or vice versa as denominator being a square root is always positive. They are equal in the case $a=b=0$ So, $a = b$ is implied and hence function is injective.

answered Sep 16 '17 at 09:33

john doe

1,075

score 1 · Answer 2 · answered Sep 16 '17 at 10:05

Suppose $f(a)=f(b)$; then with a simple manipulation $$ a\sqrt{b^2+1}=b\sqrt{a^2+1} $$ Squaring we get $$ a^2(b^2+1)=b^2(a^2+1) $$ and therefore $a^2=b^2$. This is probably where you got to. This can be written $|a|=|b|$.

However, from $$ \frac{a}{\sqrt{a^2+1}}=\frac{b}{\sqrt{b^2+1}} $$ it follows that

$a\ge0$ and $b\ge0$ or
$a<0$ and $b<0$.

In either case you conclude $a=b$.

Prove $f(x)=\frac{x}{\sqrt{x^2+1}}$ to be injective

2 Answers2