Bonjour,
1) f(x) = ax + b
f(ax + b) = a(ax + b) + b
f(a(ax + b) + b) = a[a(ax + b) + b] + b = a³x + a²b + ab + b
a³x = x/8 ⇒ a³ = 1/8 ⇒ a = 1/2
Et donc : 1/8 x b + 1/4 x b + 1/2 x b + b = -7/4
⇒ (1 + 2 + 4 + 8)/8 x b = -7/4
⇔ 15/8 x b = -7/4
⇔ b = -7/4 x 8/15 = -14/15
⇒ f(x) = x/2 - 14/15
2) f(x) = ax + b
f(0) = b et f(1) = a + b
[f(0) - f(1)] x [f(0) + f(1)] = 0
⇔ [b - (a + b)] x [b + (a + b)] = 0
⇔ (-a) x (a + 2b) = 0
⇒ a = 0 ou a + 2b = 0
Or : f(0) + f(1) = 6 ⇔ a + 2b = 6
⇒ a = 0
⇒ 2b = 6 ⇒ b = 3
⇒ f(x) = 0x + 3 = 3 (fonction constante donc)
On vérifie : f(0) + f(1) = 3 + 3 = 6
f(0) - f(1) = 3 - 3 = 0
et donc 0 x 6 = 0
