extreme f(x)=3x^3-3x^2-3x+10

Lösung

extrempunkte

f (x) = 3 x^{3} - 3 x^{2} - 3 x + 10

Maximum (- \frac{1}{3}, \frac{95}{9}), Minimum (1, 7)

Schritte zur Lösung

f^{'} (x) = 9 x^{2} - 6 x - 3

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{1}{3},

Absteigend

: - \frac{1}{3} < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = - \frac{1}{3}

3 x^{3} - 3 x^{2} - 3 x + 10 : \frac{95}{9}

Maximum (- \frac{1}{3}, \frac{95}{9})

Stecke

x = 1

3 x^{3} - 3 x^{2} - 3 x + 10 : 7

Minimum (1, 7)

Maximum (- \frac{1}{3}, \frac{95}{9}), Minimum (1, 7)

e x t re m e 0.5 x + 15 + \frac{5000}{x}

e x t re m e y = \frac{x \cdot ( 1 + x ) \cdot ( 1 - x )}{1 + 2 x}

e x t re m e f (x) = \frac{( x ^{2} )}{x ^{2} - 81}

e x t re m e f (x) = 3 x^{3} - 12 x^{2} + 12 x - 6, 0 \leq x \leq 4

e x t re m e f (x) = x (16 - 41 + 2 x) (\frac{41}{2} - x)