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

what is range of a function   f(x)= ((x+1)/(√(x^2 −1)))

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

Commented by bobhans last updated on 23/May/20

(√(x^2 −1)) is valid when x^2 > 1. so we have domain  as (min_f , 1) ∪ (1, inf ) . in first interval  the function goes from −1 to 0, since  limit (f,x, min_f )=−1 and the second   interval the function goes from inf down  to 1, since limit (f,x,1,+∞) = inf and   limit (f,x,inf) = 1 . so the range is   (−1,0) ∪ (1, ∞)

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

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

thank you

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

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