Bonjour
1.
f(1) = 2 ; g(2) = -1/2 ; f(-1/2) = 1/2 ; g(1/2) = -2 ; f(f(f(2/3)))) = -2 + 3 = 1
2. f(f(g(f(f(0))))) = f(f(g(f(1)))) = f(f(g(2))) = f(f(-1/2) = -1/2 + 2 = 3/2
Oui, en appliquant g,f,g,f,f,f
g(3/2) = -2/3
f(-2/3) = 1/3
g(1/3) = -3
f(f(f(-3))) = 0
3a. g(f(g(f(g(f(2)))))) = g(f(g(f(-1/3)))) = g(f(-3/2)) = 2
g(f(g(f(g(f(5)))))) = g(f(g(f(-1/6)))) = g(f(-6/5)) = 5
g(f(g(f(g(f(4)))))) = g(f(g(f(-1/5)))) = g(f(-5/4)) = 4
On pourrait conjecturer que g(f(g(f(g(f(x)))))) = x
b. g(f(g(f(g(f(x)))))) = g(f(g(f(-1/(x+1)))))
= g(f(g(1 - 1/(x+1))))
= g(f(g(x/(x+1))))
= g(f(-(x+1)/x))
= g(f(-1-1/x))
= g(-1/x) = x
