Show that $(1-\sqrt{x})/(1+\sqrt{x})$ is injective for $x>0$ I have just started my course in calculus and would need some guidance here. If possible, don't give complete solutions, since I need all training I can possibly get :)
Problem. The function $f$ is defined by $$f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}}\,,\:\: x>0\,.$$
Show that $f$ is injective in as many ways as possible.
My thoughts.
I have started by showing that $f$ is strictly increasing by showing that $f(x+\epsilon)-f(x)>0$ for any positive $\epsilon$ (we have not introduced derivatives yet, so I guess that was the best way to do it, but perhaps there are more efficient methods?). By the definition of injectivity, that implies that $f$ also is injective. Is that an acceptable solution?
I also tried to find an inverse by solving $y=f(x)$ for $x$. This gave me $$\sqrt{x}=\frac{1-y}{1+y}\Longrightarrow x=\left(\frac{1-y}{1+y}\right)^2,$$ but I am not entirely sure what I should do about the implication. Actually, I am not even sure if it's a problem.
Except this, I also wonder if there are any other ways that I can show the injectivity of $f$. 
 A: To show injectivity, you need to show that if $f(x_1)=f(x_2)$, then $x_1=x_2$. Your approach works just fine. Suppose that $f(x_1)=f(x_2)=y$. Your equation says that
$$
x_1=\left(\frac{1-y}{1+y}\right)^2
$$
and
$$
x_2=\left(\frac{1-y}{1+y}\right)^2
$$
That is, $x_1=x_2$.
A: You can write $f(x)$ as follows:
$$
f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}}=\frac{2}{1+\sqrt{x}}-1\,,\:\: x>0\,.
$$
Then if you accept that $g(x)=\sqrt x$ is injective then $f(x)$ should be injective too.
A: To show that
$f(x)$ is strictly decreasing,
just note that
$1-\sqrt{x}$
is strictly decreasing
and
$1+\sqrt{x}$
is strictly increasing.
Then prove that
if $a(x)$ is strictly decreasing
and
$b(x)$ is strictly increasing
and positive,
then
$\dfrac{a(x)}{b(x)}$
is strictly decreasing.
We want to show that
$\dfrac{a(x+h)}{b(x+h)}
< \dfrac{a(x)}{b(x)}
$.
$\begin{align}
\frac{a(x+h)}{b(x+h)}
&< \frac{a(x)}{b(x+h)} \quad \text{ (since } a(x+h) < a(x))\\
&< \frac{a(x)}{b(x)} \quad \text{ (since } b(x+h) > b(x) > 0)\\
\end{align}
$ 
That's all you need.
