[tex]f:(-1,1)\to(-1,1),\quad f(x) = \dfrac{2x}{1+x^2}[/tex]
Injectivitate:
[tex]\text{Daca }f(u) = f(y) \to u = v \Rightarrow f\,-\text{ injectiva}\\ \\ \dfrac{2u}{1+u^2} = \dfrac{2v}{1+v^2} \Rightarrow 2u(1+v^2) = 2v(1+u^2) \Rightarrow \\ \\ \Rightarrow 2u+2uv^2 = 2v+2u^2v \Rightarrow 2uv^2-2u^2v = 2v-2u \Rightarrow \\ \\ \Rightarrow uv(2v-2u) = 2v-2u \\ \\ \boxed{1}\quad u \neq v \Rightarrow uv = 1\quad (F) \\ \boxed{2}\quad u = v \Rightarrow 0 = 0\quad (A) \Rightarrow f\,-\text{ injectiva}[/tex]
Surjectivitate:
[tex]\forall y\in (-1,1),\quad \exists x\in (-1,1)\quad a.i.\quad f(x) = y\\ \\ f(x) = y \Rightarrow \dfrac{2x}{1+x^2} = y \Rightarrow y(1+x^2) = 2x \Rightarrow \\ \\ \Rightarrow yx^2-2x+y = 0\\ \\ \Delta_x \geq 0 \Rightarrow 4-4y^2 \geq 0,\quad \forall y\in (-1,1) \Rightarrow x\text{ exista }\forall y\in (-1,1) \\ \\ \Rightarrow f\,-\text{ surjectiva}[/tex]
Deoarece f este injectivă și surjectivă ⇒ f este bijectivă.
[tex]x = \dfrac{2y}{1+y^2} \Rightarrow xy^2-2y+x = 0 \\ \\ \Delta_y = 4 - 4x^2 \Rightarrow y_{1,2} = \dfrac{2\pm 2\sqrt{1-x^2}}{2x} = \dfrac{1\pm \sqrt{1-x^2}}{x} \\ \\ \\\text{Dar }y \in (-1,1):\\\\\Rightarrow f:(-1,1)\to (-1,1),\quad f^{-1}(x) = \dfrac{1-\sqrt{1-x^2}}{x}[/tex]
