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Question Number 130362 by greg_ed last updated on 24/Jan/21

$$\mathrm{f}\left(x\right)=\frac{2x+1}{\sqrt{x^2-\mid 2x-3\mid }}$$$$\mathrm{Domain}{\mathrm{D}}_{\mathrm{f}}=?$$

Answered by ajfour last updated on 24/Jan/21

$$x^2>\mid 2x-3\mid$$$$letx⩾\frac{3}{2}$$$$\mathrm{\&}{x}^2-2x+3>0$$$$\Rightarrow {(x-1)}^2+2>0$$$$\Rightarrow x\in [\frac{3}{2},\mathrm{\infty })$$$$andifx<\frac{3}{2}$$$$x^2+2x-3>0$$$${(x+1)}^2>4$$$$x<-3orx>1$$$$\Rightarrow x\in (-\mathrm{\infty },-3)\cup (1,\frac{3}{2})$$$$\Rightarrow x\in \mathbb{R}-[-3,1]$$

Answered by mathmax by abdo last updated on 24/Jan/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{2\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2-\mid 2\mathrm{x}-3\mid }}$$$$\mathrm{x}-\mathrm{\infty }\frac{3}{2}+\mathrm{\infty }$$$$\mid 2\mathrm{x}-3\mid -2\mathrm{x}+302\mathrm{x}-3$$$$\mathrm{f}\left(\mathrm{x}\right)\frac{2\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}-3}}...\frac{2\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2-2\mathrm{x}+3}}$$$$\mathrm{case}1\mathrm{x}<\frac{3}{2}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{2\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}-3}}$$$$\mathrm{x}\in {\mathrm{D}}_{\mathrm{f}}\iff {\mathrm{x}}^2+2\mathrm{x}-3>0\iff {\mathrm{x}}^2+2\mathrm{x}+1-4>0\Rightarrow {(\mathrm{x}+1)}^2-4>0\Rightarrow$$$$(\mathrm{x}+1-2)(\mathrm{x}+1+2)>0\Rightarrow (\mathrm{x}-1)(\mathrm{x}+3)>0\Rightarrow \mathrm{x}\in ]-\mathrm{\infty },-3[\cup ]1,+\mathrm{\infty }[\Rightarrow$$$${\mathrm{D}}_{\mathrm{f}}=]-\mathrm{\infty },-3[\cup ]1,\frac{3}{2}[$$$$\mathrm{case}2\mathrm{x}⩾\frac{3}{2}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{2\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2-2\mathrm{x}+3}}$$$$\mathrm{x}\in {\mathrm{D}}_{\mathrm{f}}\iff {\mathrm{x}}^2-2\mathrm{x}+3>0\Rightarrow {(\mathrm{x}-1)}^2+2>0\Rightarrow {\mathrm{D}}_{\mathrm{f}}=[\frac{3}{2},+\mathrm{\infty }[$$

Answered by MJS_new last updated on 25/Jan/21

$$f\left(x\right)=\frac{2x+1}{\sqrt{x^2-\mid 2x-3\mid }}$$$$\left(1\right)2x+1=0\iff x=-\frac{1}{2}\Rightarrow f\left(x\right)=0$$$$\left(2\right)x^2-\mid 2x-3\mid >0\Rightarrow x<-3\vee 1<x$$$$\Rightarrow$$$$D_f=\{x\in \mathbb{R}\mid x<-3\vee 1<x\vee x=-\frac{1}{2}\}$$

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