f(x) = ax3 + bx = x·(a·x2 + b)
f'(x) = 3ax2 + b
F(x) = 1/4·a·x4 + 1/2·b·x2
f'(√3) = 9a + b = 0
F(√(-b/a)) = 1/4·a·(-b/a)2 + 1/2·b·(-b/a) = - b2/(4·a) = 9/4
Löse das Gleichungssystem. Ich erhalte a = - 1/9 ∧ b = 1
Also probiere mal: f(x) = - 1/9·x3 + x