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

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

Commented by MJS last updated on 05/Mar/20

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

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

$$gof\left(x\right)=g\left(f\left(x\right)\right)=\frac{1}{\sqrt{{\left(f\left(x\right)\right)}^2-3}}=\frac{1}{\sqrt{x^2-1-3}}=\frac{1}{\sqrt{x^2-4}}$$$$x^2-4>0\Rightarrow \mid x\mid >2\Rightarrow x>2orx<-2\Rightarrow D_{gof}=]-\mathrm{\infty },-2[\cup ]2,+\mathrm{\infty }[$$

Answered by MJS last updated on 05/Mar/20

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

Commented by jagoll last updated on 05/Mar/20

$$\mathrm{yess}$$

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