Sagot :
Réponse :
Bonjour,
1) a) ƒ(x) = –7
⇔ 5x + 1 = –7
⇔ 5x = –8
⇔ x = –8/5
ƒ(x) = 0
⇔ 5x + 1 = 0
⇔ 5x = –1
⇔ x = –1/5
b) g(x) = –3
⇔ –2x – 4 = –3
⇔ –2x = 1
⇔ x = –1/2
g(x) = 3
⇔ –2x – 4 = 3
⇔ –2x = 7
⇔ x = –7/2
c) ƒ(1) = 5 × 1 + 1
ƒ(1) = 5 + 1
ƒ(1) = 6
g(1) = –2 × 1 – 4
g(1) = –2 – 4
g(1) = –6
d) On sait que ƒ(1) = 6 et que g(1) = –6
Or ƒ(1) ≠ g(1) ⇔ 6 ≠ –6
Donc 1 n'est pas solution de l'équation ƒ(x) = g(x)
e) ƒ(x) = g(x)
⇔ 5x + 1 = –2x – 4
⇔ 5x + 2x = –4 – 1
⇔ 7x = –5
⇔ x = –5/7