Question

prove that f(x) =x/(x squire+1),x is belongs to set of real number, is bijection.

Answer 1

Answer on Question #53953, Math Real Analysis

prove that $f(x) = x / (x \text{ squire} + 1), x$ is belongs to set of real numbers, is bijection.

Solution:

$$f (x) = \frac {x}{x ^ {2} + 1}$$

If $x\in \mathbb{R}\to f(x)\in \mathbb{R}$

$$\forall x _ {1} \in X, \forall x _ {2} \in X \Rightarrow f (x _ {1}) = \frac {x _ {1}}{x _ {1} ^ {2} + 1}; f (x _ {2}) = \frac {x _ {2}}{x _ {2} ^ {2} + 1}$$ $$\frac {x _ {1}}{x _ {1} ^ {2} + 1} = \frac {x _ {2}}{x _ {2} ^ {2} + 1} \Rightarrow (x _ {2} ^ {2} + 1) x _ {1} = (x _ {1} ^ {2} + 1) x _ {2} \Rightarrow$$ $$x _ {2} ^ {2} x _ {1} - \left(x _ {1} ^ {2} + 1\right) x _ {2} + x _ {1} = 0$$ $$D = \left(x _ {1} ^ {2} + 1\right) ^ {2} - 4 x _ {1} ^ {2} = \left(x _ {1} ^ {2} - 1\right) ^ {2}$$ $$x _ {2} = \frac {\left(x _ {1} ^ {2} + 1\right) \pm \left| x _ {1} ^ {2} - 1 \right|}{2 x _ {1}}$$

If $x_{1}^{2} - 1 > 0 \Rightarrow$

$$x _ {2 _ {1, 2}} = \frac {\left(x _ {1} ^ {2} + 1\right) \pm \left(x _ {1} ^ {2} - 1\right)}{2 x _ {1}}$$ $$x _ {2 _ {1}} = x _ {1}$$ $$x _ {2 _ {2}} = 1 / x _ {1}$$

If $x_{1}^{2} - 1 < 0 \Rightarrow$

$$x _ {2 _ {1, 2}} = \frac {\left(x _ {1} ^ {2} + 1\right) \pm \left(- x _ {1} ^ {2} + 1\right)}{2 x _ {1}}$$ $$x _ {2 _ {1}} = 1 / x _ {1}$$ $$x _ {2 _ {2}} = x _ {1}$$

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