Bonsoir,
1) g(x) = 1/3x³ - 2x² + 3x + 1
⇒ g'(x) = 1/3 × 3 × x - 2 × 2 × x + 3 = x² - 4x + 3
on cherche g'(x) = 0
Δ = b² - 4ac = 16 - 12 = 4
x1 = (-b - √Δ)/2a = (4 - 2)/2 = 1
x2 = (-b + √Δ)/2a = (4 + 2)/2 = 3
2) Tableau de variations
x | -∞ 1 3 +∞
g'(x) | + 0 - 0 +
g(x) | ↑ ↓ ↑
