Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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,+∞[

$$\left.{f}\left({x}\right)=\frac{{x}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\Rightarrow{f}\:{defined}\:{on}\right]−\infty,−\mathrm{1}\left[\cup\right]\mathrm{1},+\infty\left[\right. \\ $$$${lim}_{{x}\rightarrow−\infty} \:\:\:{f}\left({x}\right)\:={lim}_{{x}\rightarrow−\infty} \:\:\:\frac{{x}}{\mid{x}\mid\sqrt{\mathrm{1}−{x}^{−\mathrm{2}} }}\:=−\mathrm{1} \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)={lim}_{{x}\rightarrow+\infty} \:\:\:\frac{{x}}{{x}\sqrt{\mathrm{1}−{x}^{−\mathrm{2}} }}\:=\mathrm{1} \\ $$$${f}^{'} \left({x}\right)=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−{x}×\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}}{{x}^{\mathrm{2}} −\mathrm{1}}\:=\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:=\frac{−\mathrm{2}}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\frac{−\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}<\mathrm{0}\:\Rightarrow{f}\:{is}\:{decreasing}\:{on}\:{D}_{{f}} \\ $$$${x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${f}^{'} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:− \\ $$$${f}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:\:\:{decr}−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty\:{decr}\:\mathrm{1} \\ $$$$\left.\Rightarrow{f}\left(\right]−\infty,−\mathrm{1}\left[\right)\:=\right]−\infty,−\mathrm{1}\left[\:\:{and}\:{f}\left(\right]\mathrm{1},+\infty\left[\right)=\left[\mathrm{1},+\infty\left[\right.\right.\right. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com