Question Number 170522 by leicianocosta last updated on 25/May/22
Answered by FelipeLz last updated on 26/May/22
![1) f: A→B ∧ g: B→C ⇒ g○f: A→C g○f = g(f(x)) = 2f(x)+1 = 2x−1 P = {(a, c) ∈ A×C ∣ c = g(f(a))} A = {1, 2, 3} ⇒ { ((g(f(1)) = 1)),((g(f(2)) = 3)),((g(f(3)) = 5)) :} P = {(1, 1), (2, 3), (3, 5)} 2) f(x) = x^3 g(x) = x+1 g○f = g(f(x)) = f(x)+1 = x^3 +1 f○g = f(g(x)) = [g(x)]^3 = x^3 +3x^2 +3x+1 f○f = f(f(x)) = [f(x)]^3 = x^9 g○g = g(g(x)) = g(x)+1 = x+2 3) f(x) = { ((x, quando x ≤ 1)),((((x+1)/2), quando 1 < x ≤ 3)),((x^2 −7, quando x > 3)) :} g(x) = x = y x = y ∴ g^(−1) (x) = g(x) g(1) = 1 h(x) = ((x+1)/2) = y x = 2y−1 ∴ h^(−1) (x) = 2x−1 h(1) = 1, h(3) = 2 p(x) = x^2 −7 = y x = (√(y+7)) ∴ p^(−1) (x) = (√(x+7)) p(3) = 2 f^(−1) (x) = { ((x, quando x ≤ 1)),((2x−1, quando 1 < x ≤ 2)),(((√(x+7)), quando x > 2)) :}](Q170534.png)
$$\left.\mathrm{1}\right) \\ $$$$\:\:\:\:{f}:\:{A}\rightarrow{B}\:\wedge\:{g}:\:{B}\rightarrow{C}\:\Rightarrow\:{g}\circ{f}:\:{A}\rightarrow{C} \\ $$$$\:\:\:\:{g}\circ{f}\:=\:{g}\left({f}\left({x}\right)\right)\:=\:\mathrm{2}{f}\left({x}\right)+\mathrm{1}\:=\:\mathrm{2}{x}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathbb{P}\:=\:\left\{\left({a},\:{c}\right)\:\in\:{A}×{C}\:\mid\:{c}\:=\:{g}\left({f}\left({a}\right)\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{A}\:=\:\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3}\right\}\:\Rightarrow\:\begin{cases}{{g}\left({f}\left(\mathrm{1}\right)\right)\:=\:\mathrm{1}}\\{{g}\left({f}\left(\mathrm{2}\right)\right)\:=\:\mathrm{3}}\\{{g}\left({f}\left(\mathrm{3}\right)\right)\:=\:\mathrm{5}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathbb{P}\:=\:\left\{\left(\mathrm{1},\:\mathrm{1}\right),\:\left(\mathrm{2},\:\mathrm{3}\right),\:\left(\mathrm{3},\:\mathrm{5}\right)\right\} \\ $$$$\left.\mathrm{2}\right)\: \\ $$$$\:\:\:\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \\ $$$$\:\:\:\:{g}\left({x}\right)\:=\:{x}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{g}\circ{f}\:=\:{g}\left({f}\left({x}\right)\right)\:=\:{f}\left({x}\right)+\mathrm{1}\:=\:{x}^{\mathrm{3}} +\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\circ{g}\:=\:{f}\left({g}\left({x}\right)\right)\:=\:\left[{g}\left({x}\right)\right]^{\mathrm{3}} \:=\:{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\circ{f}\:=\:{f}\left({f}\left({x}\right)\right)\:=\:\left[{f}\left({x}\right)\right]^{\mathrm{3}} \:=\:{x}^{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{g}\circ{g}\:=\:{g}\left({g}\left({x}\right)\right)\:=\:{g}\left({x}\right)+\mathrm{1}\:=\:{x}+\mathrm{2} \\ $$$$\left.\mathrm{3}\right) \\ $$$$\:\:\:\:\:{f}\left({x}\right)\:=\:\begin{cases}{{x},\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{quando}\:{x}\:\leqslant\:\mathrm{1}}\\{\frac{{x}+\mathrm{1}}{\mathrm{2}},\:\:\:\:\:\:\:\mathrm{quando}\:\mathrm{1}\:<\:{x}\:\leqslant\:\mathrm{3}}\\{{x}^{\mathrm{2}} −\mathrm{7},\:\:\:\:\:\mathrm{quando}\:{x}\:>\:\mathrm{3}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{g}\left({x}\right)\:=\:{x}\:=\:{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:=\:{y}\:\therefore\:{g}^{−\mathrm{1}} \left({x}\right)\:=\:{g}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{g}\left(\mathrm{1}\right)\:=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{h}\left({x}\right)\:=\:\frac{{x}+\mathrm{1}}{\mathrm{2}}\:=\:{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:=\:\mathrm{2}{y}−\mathrm{1}\:\therefore\:{h}^{−\mathrm{1}} \left({x}\right)\:=\:\mathrm{2}{x}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{h}\left(\mathrm{1}\right)\:=\:\mathrm{1},\:\:{h}\left(\mathrm{3}\right)\:=\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{p}\left({x}\right)\:=\:{x}^{\mathrm{2}} −\mathrm{7}\:=\:{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:=\:\sqrt{{y}+\mathrm{7}}\:\therefore\:{p}^{−\mathrm{1}} \left({x}\right)\:=\:\sqrt{{x}+\mathrm{7}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{p}\left(\mathrm{3}\right)\:=\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:{f}^{−\mathrm{1}} \left({x}\right)\:=\:\begin{cases}{{x},\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{quando}\:{x}\:\leqslant\:\mathrm{1}}\\{\mathrm{2}{x}−\mathrm{1},\:\:\:\:\:\:\mathrm{quando}\:\mathrm{1}\:<\:{x}\:\leqslant\:\mathrm{2}}\\{\sqrt{{x}+\mathrm{7}},\:\:\:\:\:\:\:\mathrm{quando}\:{x}\:>\:\mathrm{2}}\end{cases} \\ $$