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Question Number 95140 by i jagooll last updated on 23/May/20

$$\mathrm{what}\mathrm{is}\mathrm{range}\mathrm{of}\mathrm{a}\mathrm{function}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2-1}}$$

Commented by bobhans last updated on 23/May/20

$$\sqrt{{\mathrm{x}}^2-1}\mathrm{is}\mathrm{valid}\mathrm{when}{\mathrm{x}}^2>1.\mathrm{so}\mathrm{we}\mathrm{have}\mathrm{domain}$$$$\mathrm{as}({\min }_{\mathrm{f}},1)\cup (1,\inf ).\mathrm{in}\mathrm{first}\mathrm{interval}$$$$\mathrm{the}\mathrm{function}\mathrm{goes}\mathrm{from}-1\mathrm{to}0,\mathrm{since}$$$$\mathrm{limit}(\mathrm{f},\mathrm{x},{\min }_{\mathrm{f}})=-1\mathrm{and}\mathrm{the}\mathrm{second}$$$$\mathrm{interval}\mathrm{the}\mathrm{function}\mathrm{goes}\mathrm{from}\inf \mathrm{down}$$$$\mathrm{to}1,\mathrm{since}\mathrm{limit}(\mathrm{f},\mathrm{x},1,+\mathrm{\infty })=\inf \mathrm{and}$$$$\mathrm{limit}(\mathrm{f},\mathrm{x},\inf )=1.\mathrm{so}\mathrm{the}\mathrm{range}\mathrm{is}$$$$(-1,0)\cup (1,\mathrm{\infty })$$$$$$

Commented by i jagooll last updated on 23/May/20

$$\mathrm{thank}\mathrm{you}$$

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