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Question Number 83539 by jagoll last updated on 03/Mar/20
what is range of function   f(x) = (x/( (√(x^2 −1))))?
$$\mathrm{what}\:\mathrm{is}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}? \\ $$
Commented by john santu last updated on 03/Mar/20
domain x^2 −1>0 ⇒ x<−1 ∨x >1  y(√(x^2 −1)) = x ⇒ (x^2 −1)y^2 =x^2   (y^2 −1)x^2 −y^2  = 0  Δ = 4y^2 (y^2 −1)≥0  ⇒ y^2 (y−1)(y+1)≥0  Range ⇒ y≤−1∨ y≥1
$$\mathrm{domain}\:\mathrm{x}^{\mathrm{2}} −\mathrm{1}>\mathrm{0}\:\Rightarrow\:\mathrm{x}<−\mathrm{1}\:\vee\mathrm{x}\:>\mathrm{1} \\ $$$$\mathrm{y}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:=\:\mathrm{x}\:\Rightarrow\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{y}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} \\ $$$$\left(\mathrm{y}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Delta\:=\:\mathrm{4y}^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} −\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{y}^{\mathrm{2}} \left(\mathrm{y}−\mathrm{1}\right)\left(\mathrm{y}+\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{Range}\:\Rightarrow\:\mathrm{y}\leqslant−\mathrm{1}\vee\:\mathrm{y}\geqslant\mathrm{1}\: \\ $$
Commented by john santu last updated on 03/Mar/20
Commented by mathmax by abdo last updated on 03/Mar/20
f(x)=(x/( (√(x^2 −1)))) ⇒f defined on]−∞,−1[∪]1,+∞[  lim_(x→−∞)    f(x) =lim_(x→−∞)    (x/(∣x∣(√(1−x^(−2) )))) =−1  lim_(x→+∞) f(x)=lim_(x→+∞)    (x/(x(√(1−x^(−2) )))) =1  f^′ (x)=(((√(x^2 −1))−x×((2x)/(2(√(x^2 −1)))))/(x^2 −1)) =((2(x^2 −1)−2x^2 )/(2(x^2 −1)(√(x^2 −1)))) =((−2)/(2(x^2 −1)(√(x^2 −1))))  =((−1)/((x^2 −1)(√(x^2 −1))))<0 ⇒f is decreasing on D_f   x                   −∞                 −1                 1                 +∞  f^′                                     −             ∣∣             ∣∣               −  f                         −1   decr−∞                    +∞ decr 1  ⇒f(]−∞,−1[) =]−∞,−1[  and f(]1,+∞[)=[1,+∞[
$$\left.{f}\left({x}\right)=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\Rightarrow{f}\:{defined}\:{on}\right]−\infty,−\mathrm{1}\left[\cup\right]\mathrm{1},+\infty\left[\right. \\ $$$${lim}_{{x}\rightarrow−\infty} \:\:\:{f}\left({x}\right)\:={lim}_{{x}\rightarrow−\infty} \:\:\:\frac{{x}}{\mid{x}\mid\sqrt{\mathrm{1}−{x}^{−\mathrm{2}} }}\:=−\mathrm{1} \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)={lim}_{{x}\rightarrow+\infty} \:\:\:\frac{{x}}{{x}\sqrt{\mathrm{1}−{x}^{−\mathrm{2}} }}\:=\mathrm{1} \\ $$$${f}^{'} \left({x}\right)=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−{x}×\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}}{{x}^{\mathrm{2}} −\mathrm{1}}\:=\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:=\frac{−\mathrm{2}}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\frac{−\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}<\mathrm{0}\:\Rightarrow{f}\:{is}\:{decreasing}\:{on}\:{D}_{{f}} \\ $$$${x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${f}^{'} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:− \\ $$$${f}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:\:\:{decr}−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty\:{decr}\:\mathrm{1} \\ $$$$\left.\Rightarrow{f}\left(\right]−\infty,−\mathrm{1}\left[\right)\:=\right]−\infty,−\mathrm{1}\left[\:\:{and}\:{f}\left(\right]\mathrm{1},+\infty\left[\right)=\left[\mathrm{1},+\infty\left[\right.\right.\right. \\ $$

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