Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 31794 by mondodotto@gmail.com last updated on 14/Mar/18

Commented by MJS last updated on 14/Mar/18

$$f\left(x\right)=\sqrt{x^3}=x\sqrt{x}$$$$g\left(x\right)=\frac{1}{\sqrt{x}}=\frac{\sqrt{x}}{x}$$$$f\left(x\right)-g\left(x\right)=x\sqrt{x}-\frac{\sqrt{x}}{x}=(x-\frac{1}{x})\sqrt{x}$$$$\mathrm{domain}=\{x\in \mathbb{R}\mid x>0\}={\mathbb{R}}^{+}$$$$\mathrm{range}=\mathbb{R}$$$$\frac{f\left(x\right)}{g\left(x\right)}=x\sqrt{x}/\frac{\sqrt{x}}{x}=x\sqrt{x}\times \frac{x}{\sqrt{x}}=x^2$$$$\mathrm{it}\mathrm{looks}\mathrm{like}\mathrm{domain}=\mathbb{R}\mathrm{because}$$$$\mathrm{you}\mathrm{can}\mathrm{reduce}\mathrm{the}\mathrm{fraction}$$$$\frac{\sqrt{x}}{\sqrt{x}}\mathrm{when}x<0(\mathrm{whatever}\mathrm{i}.\mathrm{e}.$$$$\sqrt{-3}\mathrm{might}\mathrm{be},\frac{\sqrt{-3}}{\sqrt{-3}}=1)\mathrm{but}\mathrm{you}$$$${\mathrm{can}}^{\prime }\mathrm{t}\mathrm{when}x=0(\mathrm{although}\mathrm{in}\mathrm{some}$$$$\mathrm{cases}\mathrm{it}\mathrm{might}\mathrm{be}\mathrm{useful}\mathrm{to}\mathrm{add}$$$$(x,y)=(0,1)\mathrm{to}\mathrm{close}\mathrm{the}\mathrm{gap}),\mathrm{so}$$$$\mathrm{domain}=\{x\in \mathbb{R}\mid x\ne 0\}=\mathbb{R}\mathrm{\setminus }\left\{0\right\}$$$$\mathrm{range}=\{x\in \mathbb{R}\mid x>0\}={\mathbb{R}}^{+}$$

Commented by mondodotto@gmail.com last updated on 15/Mar/18

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com