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Suppose-that-f-x-1-x-1-and-g-x-4-x-1-Find-the-domain-of-each-of-the-composition-a-f-o-g-b-f-o-f-

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Question Number 10887 by Saham last updated on 28/Feb/17
Suppose that f(x) = (1/(x + 1)) and g(x) = (4/(x + 1)) . Find the domain of each of the composition (a) f o g (b) f o f
$$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{x}\:+\:\mathrm{1}}\:\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{4}}{\mathrm{x}\:+\:\mathrm{1}}\:. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{composition}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{f}\:\mathrm{o}\:\mathrm{g}\:\:\:\left(\mathrm{b}\right)\:\:\mathrm{f}\:\mathrm{o}\:\mathrm{f} \\ $$
Answered by geovane10math last updated on 28/Feb/17
I wrote an answer
$$\mathrm{I}\:\mathrm{wrote}\:\mathrm{an}\:\mathrm{answer} \\ $$
Commented by geovane10math last updated on 28/Feb/17
f(x) = (1/(x + 1)) g(x) = (4/(x + 1)) a f(g(x)) = f((4/(x + 1))) = (1/((4/(x + 1)) + 1)) Domain : (4/(x + 1)) + 1 ≠ 0 (4/(x + 1)) ≠ −1 (4/(−1)) ≠ x + 1 −4 ≠ x + 1 x ≠ −5 b f(f(x)) = f((1/(x + 1))) = (1/((1/(x + 1)) + 1)) Domain: (1/(x + 1)) + 1 ≠ 0 (1/(x + 1)) ≠ −1 (1/(−1)) ≠ x + 1 −1 ≠ x + 1 x ≠ −2
$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}\:+\:\mathrm{1}}\:\:\:\:\:\:\:\:\:{g}\left({x}\right)\:=\:\frac{\mathrm{4}}{{x}\:+\:\mathrm{1}} \\ $$$${a}\:\:\:{f}\left({g}\left({x}\right)\right)\:=\:{f}\left(\frac{\mathrm{4}}{{x}\:+\:\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\frac{\mathrm{4}}{{x}\:+\:\mathrm{1}}\:+\:\mathrm{1}} \\ $$$$\mathrm{Domain}\::\:\frac{\mathrm{4}}{{x}\:+\:\mathrm{1}}\:+\:\mathrm{1}\:\neq\:\mathrm{0} \\ $$$$\frac{\mathrm{4}}{{x}\:+\:\mathrm{1}}\:\neq\:−\mathrm{1} \\ $$$$\frac{\mathrm{4}}{−\mathrm{1}}\:\neq\:{x}\:+\:\mathrm{1} \\ $$$$−\mathrm{4}\:\neq\:{x}\:+\:\mathrm{1} \\ $$$$\boldsymbol{{x}}\:\neq\:−\mathrm{5} \\ $$$${b}\:\:{f}\left({f}\left({x}\right)\right)\:=\:{f}\left(\frac{\mathrm{1}}{{x}\:+\:\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{{x}\:+\:\mathrm{1}}\:+\:\mathrm{1}} \\ $$$$\mathrm{Domain}:\:\frac{\mathrm{1}}{{x}\:+\:\mathrm{1}}\:+\:\mathrm{1}\:\neq\:\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{x}\:+\:\mathrm{1}}\:\neq\:−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{−\mathrm{1}}\:\neq\:{x}\:+\:\mathrm{1} \\ $$$$−\mathrm{1}\:\neq\:\:{x}\:+\:\mathrm{1} \\ $$$$\boldsymbol{{x}}\:\neq\:−\mathrm{2} \\ $$$$ \\ $$
Commented by Saham last updated on 28/Feb/17
i did not see it sir.
$$\mathrm{i}\:\mathrm{did}\:\mathrm{not}\:\mathrm{see}\:\mathrm{it}\:\mathrm{sir}. \\ $$
Commented by Saham last updated on 28/Feb/17
I really appreciate your effort sir. God bless you.
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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