Question-206764

Question Number 206764 by mustafazaheen last updated on 24/Apr/24

Answered by A5T last updated on 24/Apr/24

f(g(−3))=f((√(−3)))=f((√3)i)=((√3)i)^2 =−3

$${f}\left({g}\left(−\mathrm{3}\right)\right)={f}\left(\sqrt{−\mathrm{3}}\right)={f}\left(\sqrt{\mathrm{3}}{i}\right)=\left(\sqrt{\mathrm{3}}{i}\right)^{\mathrm{2}} =−\mathrm{3} \\ $$

Commented by JDamian last updated on 24/Apr/24

(√(−3 ))≠ 3i (√(−3)) = (√3)i ⇒ ((√3)i)^2 = −3 because g(x) = f^(−1) (x) ⇒ f○g = g ○f = x

$$\sqrt{−\mathrm{3}\:}\neq\:\mathrm{3}{i} \\ $$$$\sqrt{−\mathrm{3}}\:=\:\sqrt{\mathrm{3}}{i}\:\:\Rightarrow\:\left(\sqrt{\mathrm{3}}{i}\right)^{\mathrm{2}} \:=\:−\mathrm{3} \\ $$$$ \\ $$$${because}\:\:{g}\left({x}\right)\:=\:{f}^{−\mathrm{1}} \left({x}\right)\:\Rightarrow\:{f}\circ{g}\:=\:{g}\:\circ{f}\:=\:{x} \\ $$

Commented by mr W last updated on 24/Apr/24

for x∈R f_1 (x)=((√x))^2 is valid for x≥0 f_2 (x)=x f_1 (−3)≠−3, because f_1 (−3) is not defined. f_2 (−3)=−3

$${for}\:{x}\in{R} \\ $$$${f}_{\mathrm{1}} \left({x}\right)=\left(\sqrt{{x}}\right)^{\mathrm{2}} \:{is}\:{valid}\:{for}\:{x}\geqslant\mathrm{0} \\ $$$${f}_{\mathrm{2}} \left({x}\right)={x} \\ $$$${f}_{\mathrm{1}} \left(−\mathrm{3}\right)\neq−\mathrm{3},\:{because}\:{f}_{\mathrm{1}} \left(−\mathrm{3}\right)\:{is}\:{not}\:{defined}.\: \\ $$$${f}_{\mathrm{2}} \left(−\mathrm{3}\right)=−\mathrm{3} \\ $$

Commented by A5T last updated on 24/Apr/24

$${Yea},\:{it}\:{depends}\:{on}\:{the}\:{domain}. \\ $$

Answered by A5T last updated on 24/Apr/24

$${f}\left({g}\left({x}\right)\right)=\left(\sqrt{{x}}\right)^{\mathrm{2}} ={x}\Rightarrow{f}\left({g}\left(−\mathrm{3}\right)\right)=−\mathrm{3} \\ $$

Answered by aba last updated on 29/Apr/24

$$−\mathrm{3} \\ $$

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