Question Number 83662 by jagoll last updated on 05/Mar/20
$$\mathrm{if}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{3}}} \\ $$$$\mathrm{find}\:\mathrm{domain}\:\mathrm{function}\: \\ $$$$\left(\mathrm{g}\:\bullet\:\mathrm{f}\right)\left(\mathrm{x}\right) \\ $$
Commented by MJS last updated on 05/Mar/20
$${f}\left({g}\left({x}\right)\right)\:\mathrm{or}\:{g}\left({f}\left({x}\right)\right)\:? \\ $$$$\mathrm{both}\:\mathrm{interpretations}\:\mathrm{of}\:\left({g}\circ{f}\right)\:\mathrm{are}\:\mathrm{common} \\ $$
Commented by jagoll last updated on 05/Mar/20
$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{g}\left(\mathrm{f}\left({x}\right)\right) \\ $$
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,+∞[](https://www.tinkutara.com/question/Q83770.png)
$${gof}\left({x}\right)={g}\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{\:\sqrt{\left({f}\left({x}\right)\right)^{\mathrm{2}} −\mathrm{3}}}\:=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}−\mathrm{3}}}\:=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}} \\ $$$$\left.{x}^{\mathrm{2}} −\mathrm{4}\:>\mathrm{0}\:\Rightarrow\mid{x}\mid>\mathrm{2}\:\Rightarrow{x}>\mathrm{2}\:{or}\:{x}<−\mathrm{2}\:\Rightarrow{D}_{{gof}} =\right]−\infty,−\mathrm{2}\left[\cup\right]\mathrm{2},+\infty\left[\right. \\ $$
Answered by MJS last updated on 05/Mar/20
$${f}\left({g}\left({x}\right)\right)=\sqrt{\frac{\mathrm{4}−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{3}}} \\ $$$$\mathrm{3}<{x}^{\mathrm{2}} \leqslant\mathrm{4}\:\Rightarrow\:−\mathrm{2}\leqslant{x}<−\sqrt{\mathrm{3}}\:\vee\:\sqrt{\mathrm{3}}<{x}\leqslant\mathrm{2} \\ $$$${g}\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}>\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{2}} >\mathrm{4}\:\Rightarrow\:{x}<−\mathrm{2}\vee{x}>\mathrm{2} \\ $$
Commented by jagoll last updated on 05/Mar/20
$$\mathrm{yess} \\ $$