Define the function f: $(2,\infty) -> (-\infty,-1)$ by $f(x)= \frac{-x}{x-2}$. Show that f is bijective.
I know i need to prove both injective and surjective, and I was able to solve the equation for $x=\frac{2y}{y+1}$ , but I dont know where to take it from there.
On
Injective: $$f_1(x_1)=f_2(x_2) \Rightarrow x_1 = x_2$$ Surjective: $$y \in \mathbb{R} \Rightarrow \exists x \in \mathbb{R}, y=f_1(x)$$ In you case: $$$$ Injective: $$\frac{-x_1}{x_1-2}=\frac{-x_2}{x_2-2} \Rightarrow x_1=x_2$$ Surjective: $$y=f_1(x) \Leftrightarrow y=\frac{-x}{x-2} \Leftrightarrow x=\frac{2y}{1+y}$$ Since the function is both injective and surjective it is bijective. $\blacksquare$
Let $a, b\in (2, \infty)$. Suppose $f(a) = f(b)$. Then $\dfrac{a}{a-2} = \dfrac{b}{b-2}$. Solving yields $ab - 2a = ab - 2b$, which yields $a = b$. So you have injectivity.
Now for surjectivity, take $y \in (-\infty, -1)$. Is there an $x \in (2, \infty)$ such that $f(x) = y$? If you can solve for such $x$, then you have an answer of "yes," and so surjectivity follows.