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Question Number 75527 by liki last updated on 12/Dec/19

Commented by liki last updated on 12/Dec/19

$. . . e m e r g e n c y p l z i n e e d s o m e o n e t o c h e c k$
	Answered by MJS last updated on 12/Dec/19

	$y = \sqrt{2 x^{2} - 1}$ $minimum at x = \pm \frac{\sqrt{2}}{2} \Rightarrow y = 0$ $maximum {doesn}^{'} t exist because$ $\lim_{x \to \pm \infty} \sqrt{2 x^{2} - 1} = + \infty$ $\Rightarrow range is [0; + \infty [$
	Commented by liki last updated on 12/Dec/19

	$. . G o d b l e s s y o u s i r, i a p p r i c i a t e y o u t i m e .$
	Commented by liki last updated on 12/Dec/19

	$. . . t h a n k y o u s o m u c h s i r .$
	Commented by MJS last updated on 12/Dec/19
	$defined for x∈R\]−((√2)/2); ((√2)/2)[$
	$defined for x \in R ∖] - \frac{\sqrt{2}}{2}; \frac{\sqrt{2}}{2} [$
	Commented by liki last updated on 12/Dec/19

	$. . . s o r y s i r h o w a b o u t i f i s a y m a k e x t h e s u b j e c t ?$
	Commented by MJS last updated on 12/Dec/19

	$y = \sqrt{2 x^{2} - 1}$ $y^{2} = 2 x^{2} - 1$ $x^{2} = \frac{y^{2} + 1}{2}$ $x = \pm \frac{\sqrt{2}}{2} \sqrt{y^{2} + 1}$ $but to ensure {it}^{'} s really the inverse of$ $y = \sqrt{2 x^{2} - 1} we must keep y ⩾ 0$ $y ⩾ 0 \Rightarrow (x < - \frac{\sqrt{2}}{2} \lor \frac{\sqrt{2}}{2} < x)$
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