Réponse :
Bonjour
1) f(0) = 3
f'(0) = 1
f'(1) = 0
2)
a) f(x) = (ax + b)eˣ + c
f'(x) = aeˣ + (ax + b)eˣ = (a + ax + b)eˣ
b) On sait que f(0) = 3
⇔ (a×0 + b)e⁰ + c = 3
⇔ b + c = 3
On sait que f'(0) = 1
⇔ (a×0 + a + b)e⁰ = 1
⇔ a + b = 1
On sait que f'(1) = 0
⇔ (a×1 + a + b)e¹ = 0
⇔ (2a + b)e = 0
⇔ 2a + b = 0
c) on a donc le système suivant
b + c = 3 b + c = 3 2/3 + c = 3 c = 7/3
a + b = 1 ⇔ a + 2a = 1 ⇔ a = 1/3 ⇔ a = 1/3
2a + b = 0 b = 2a b = 2/3 b = 2/3
Au final , on a f(x) = (x/3 + 2/3)eˣ + 7/3
