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

4.Let f(x)=x^2 and g(x)=(√(x )) then find fog(x) and gof(x) and domain of(fog)(x)and(gof) are they they the same?explain. please help me???

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

Answered by amin96 last updated on 23/Aug/21

fog(x)=f(g(x))=((√x))^2 =x gof(x)=g(f(x))=(√((x)^2 ))=x fog(x)=gof(x)

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

Commented by MJS_new last updated on 23/Aug/21

you are wrong f(g(−1)) is not defined ⇒ f(g(x))≠x g(f(−1))=1≠−1 ⇒ g(f(x))≠x f(g(x))≠g(f(x)) ∀x<0

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

Answered by MJS_new last updated on 23/Aug/21

f(x)=x^2 is defined for x∈R g(x)=(√x) is defined for x∈R_0 ^+ f(g(x))=((√x))^2 =x with x∈R_0 ^+ g(f(x))=(√x^2 )=∣x∣ with x∈R

$${f}\left({x}\right)={x}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R} \\ $$$${g}\left({x}\right)=\sqrt{{x}}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}_{\mathrm{0}} ^{+} \\ $$$${f}\left({g}\left({x}\right)\right)=\left(\sqrt{{x}}\right)^{\mathrm{2}} ={x}\:\mathrm{with}\:{x}\in\mathbb{R}_{\mathrm{0}} ^{+} \\ $$$${g}\left({f}\left({x}\right)\right)=\sqrt{{x}^{\mathrm{2}} }=\mid{x}\mid\:\mathrm{with}\:{x}\in\mathbb{R} \\ $$

Commented by abdurehime last updated on 23/Aug/21

so are they the same or not??

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

Commented by MJS_new last updated on 23/Aug/21

not. f(g(x)) is not defined for x<0 g(f(x)) is defined for x<0

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

Commented by abdurehime last updated on 23/Aug/21

thanks let try to question 1 please

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

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

nice!

$$\mathrm{nice}! \\ $$

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