Question Number 75527 by liki last updated on 12/Dec/19
Commented by liki last updated on 12/Dec/19
$$...{emergency}\:{plz}\:{i}\:{need}\:{someone}\:{to}\:{check}\: \\ $$
Answered by MJS last updated on 12/Dec/19
$${y}=\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{minimum}\:\mathrm{at}\:{x}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\:{y}=\mathrm{0} \\ $$$$\mathrm{maximum}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist}\:\mathrm{because} \\ $$$$\underset{{x}\rightarrow\pm\infty} {\mathrm{lim}}\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}=+\infty \\ $$$$\Rightarrow\:\mathrm{range}\:\mathrm{is}\:\left[\mathrm{0};\:+\infty\left[\right.\right. \\ $$
Commented by liki last updated on 12/Dec/19
$$..{God}\:{bless}\:{you}\:{sir},{i}\:{appriciate}\:{you}\:{time}. \\ $$
Commented by liki last updated on 12/Dec/19
$$...{thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$
Commented by MJS last updated on 12/Dec/19
![defined for x∈R\]−((√2)/2); ((√2)/2)[](Q75533.png)
$$\left.\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\right]−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}};\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left[\right. \\ $$
Commented by liki last updated on 12/Dec/19
$$...{sory}\:{sir}\:{how}\:{about}\:{if}\:{i}\:{say}\:{make}\:{x}\:{the}\:{subject}? \\ $$
Commented by MJS last updated on 12/Dec/19
$${y}=\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${y}^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\frac{{y}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{but}\:\mathrm{to}\:\mathrm{ensure}\:\mathrm{it}'\mathrm{s}\:\mathrm{really}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of} \\ $$$${y}=\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{we}\:\mathrm{must}\:\mathrm{keep}\:{y}\geqslant\mathrm{0} \\ $$$${y}\geqslant\mathrm{0}\:\Rightarrow\:\left({x}<−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\vee\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}<{x}\right) \\ $$