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Question Number 83662 by jagoll last updated on 05/Mar/20

if f(x) = (√(x^2 −1)) and g(x) = (1/(√(x^2 −3))) find domain function (g • f)(x)

$$\mathrm{if}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{3}}} \\ $$$$\mathrm{find}\:\mathrm{domain}\:\mathrm{function}\: \\ $$$$\left(\mathrm{g}\:\bullet\:\mathrm{f}\right)\left(\mathrm{x}\right) \\ $$

Commented by MJS last updated on 05/Mar/20

f(g(x)) or g(f(x)) ? both interpretations of (g○f) are common

$${f}\left({g}\left({x}\right)\right)\:\mathrm{or}\:{g}\left({f}\left({x}\right)\right)\:? \\ $$$$\mathrm{both}\:\mathrm{interpretations}\:\mathrm{of}\:\left({g}\circ{f}\right)\:\mathrm{are}\:\mathrm{common} \\ $$

Commented by jagoll last updated on 05/Mar/20

yes sir g(f(x))

$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{g}\left(\mathrm{f}\left({x}\right)\right) \\ $$

Commented by mathmax by abdo last updated on 05/Mar/20

gof(x)=g(f(x))=(1/(√((f(x))^2 −3))) =(1/(√(x^2 −1−3))) =(1/(√(x^2 −4))) x^2 −4 >0 ⇒∣x∣>2 ⇒x>2 or x<−2 ⇒D_(gof) =]−∞,−2[∪]2,+∞[

$${gof}\left({x}\right)={g}\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{\sqrt{\left({f}\left({x}\right)\right)^{\mathrm{2}} −\mathrm{3}}}\:=\frac{\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}−\mathrm{3}}}\:=\frac{\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}} \\ $$$$\left.{x}^{\mathrm{2}} −\mathrm{4}\:>\mathrm{0}\:\Rightarrow\mid{x}\mid>\mathrm{2}\:\Rightarrow{x}>\mathrm{2}\:{or}\:{x}<−\mathrm{2}\:\Rightarrow{D}_{{gof}} =\right]−\infty,−\mathrm{2}\left[\cup\right]\mathrm{2},+\infty\left[\right. \\ $$

Answered by MJS last updated on 05/Mar/20

f(g(x))=(√((4−x^2 )/(x^2 −3))) 3<x^2 ≤4 ⇒ −2≤x<−(√3) ∨ (√3)<x≤2 g(f(x))=(1/(√(x^2 −4))) x^2 −4>0 ⇒ x^2 >4 ⇒ x<−2∨x>2

$${f}\left({g}\left({x}\right)\right)=\sqrt{\frac{\mathrm{4}−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{3}}} \\ $$$$\mathrm{3}<{x}^{\mathrm{2}} \leqslant\mathrm{4}\:\Rightarrow\:−\mathrm{2}\leqslant{x}<−\sqrt{\mathrm{3}}\:\vee\:\sqrt{\mathrm{3}}<{x}\leqslant\mathrm{2} \\ $$$${g}\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}>\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{2}} >\mathrm{4}\:\Rightarrow\:{x}<−\mathrm{2}\vee{x}>\mathrm{2} \\ $$

Commented by jagoll last updated on 05/Mar/20

yess

$$\mathrm{yess} \\ $$

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