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Question Number 151804 by abdurehime last updated on 23/Aug/21

$$4.\mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2\mathrm{and}\mathrm{g}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}$$$$\mathrm{then}\mathrm{find}\mathrm{fog}\left(\mathrm{x}\right)\mathrm{and}\mathrm{gof}\left(\mathrm{x}\right)\mathrm{and}$$$$\mathrm{domain}\mathrm{of}\left(\mathrm{fog}\right)\left(\mathrm{x}\right)\mathrm{and}\left(\mathrm{gof}\right)$$$$\mathrm{are}\mathrm{they}\mathrm{they}\mathrm{the}\mathrm{same}?\mathrm{explain}.$$$$$$$$\mathrm{please}\mathrm{help}\mathrm{me}???$$$$$$

Answered by amin96 last updated on 23/Aug/21

$$fog\left(x\right)=f\left(g\left(x\right)\right)={\left(\sqrt{x}\right)}^2=x$$$$gof\left(x\right)=g\left(f\left(x\right)\right)=\sqrt{{\left(x\right)}^2}=x$$$$fog\left(x\right)=gof\left(x\right)$$

Commented by MJS_new last updated on 23/Aug/21

$$\mathrm{you}\mathrm{are}\mathrm{wrong}$$$$f\left(g(-1)\right)\mathrm{is}\mathrm{not}\mathrm{defined}\Rightarrow f\left(g\left(x\right)\right)\ne x$$$$g\left(f(-1)\right)=1\ne -1\Rightarrow g\left(f\left(x\right)\right)\ne x$$$$f\left(g\left(x\right)\right)\ne g\left(f\left(x\right)\right)\mathrm{\forall }x<0$$

Answered by MJS_new last updated on 23/Aug/21

$$f\left(x\right)=x^2\mathrm{is}\mathrm{defined}\mathrm{for}x\in \mathbb{R}$$$$g\left(x\right)=\sqrt{x}\mathrm{is}\mathrm{defined}\mathrm{for}x\in {\mathbb{R}}_0^{+}$$$$f\left(g\left(x\right)\right)={\left(\sqrt{x}\right)}^2=x\mathrm{with}x\in {\mathbb{R}}_0^{+}$$$$g\left(f\left(x\right)\right)=\sqrt{x^2}=\mid x\mid \mathrm{with}x\in \mathbb{R}$$

Commented by abdurehime last updated on 23/Aug/21

$$\mathrm{so}\mathrm{are}\mathrm{they}\mathrm{the}\mathrm{same}\mathrm{or}\mathrm{not}??$$

Commented by MJS_new last updated on 23/Aug/21

$$\mathrm{not}.$$$$f\left(g\left(x\right)\right)\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}x<0$$$$g\left(f\left(x\right)\right)\mathrm{is}\mathrm{defined}\mathrm{for}x<0$$

Commented by abdurehime last updated on 23/Aug/21

$$\mathrm{thanks}\mathrm{let}\mathrm{try}\mathrm{to}\mathrm{question}1\mathrm{please}$$

Commented by otchereabdullai@gmail.com last updated on 23/Aug/21

$$\mathrm{nice}!$$

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