zu a)
f1(x) = x3 - 9x | Null setzen
ð x3 - 9x = 0 | x ausklammern
ð x*(x2 - 9) = 0 => x = 0 und (x2 - 9) = 0 | 3. Binom => (x+3)*(x-3) = 0
ð x2: x+3 = 0 => x2 = -3 und x3: x-3 = 0 => x3 = 3
Versuche es mit den beiden anderen analog
zu b)
x3 - 9x + x*(x+3)2 + (-x2*(x+3)) = 0
x3 - 9x + x*(x2 + 6x + 9) + (-x3 – 3x2) = 0
x3 - 9x + x3 + 6x2 + 9x -x3 - 3x2 = 0
x3 + 3x2 | x2 ausklammern
x2*(x+3)=0
x1 = 0
x2 = -3
Nullstellen berechnen
f’(x) = 3x2 + 6x | :3
f’(x) =x2 + 2x | p=2; q=0
x1/2 = -p/2 +-Wurzel ((p/2)2-0)
x1/2 = -1 +- 1 => x1 = 0 und x2 = -2