Bonjour,
Question a :
f(x) = (x+1)eˣ
u(x) = x + 1 v(x) = eˣ
u'(x) = 1 v'(x) = eˣ
f'(x) = u'v + uv' = 1 × eˣ + (x+1) × eˣ = eˣ + xeˣ + eˣ = eˣ( x + 2)
Question b :
eˣ > 0
On va donc étudier le signe de x + 2 :
x + 2 > 0
x > - 2
Tableau de variation en pièce-jointe !
Question c :
y = f'(a)(x-a) + f(a)
f(0) = (0+1)e⁰ = 1 × 1 = 1
f'(0) = e⁰( 0 + 2) = 1 × 2 = 2
y = 2(x - 0) + 1
y = 2x + 1