extreme f(x)=4x^3-2x+3

Lösung

extrempunkte

f (x) = 4 x^{3} - 2 x + 3

Maximum (- \frac{1}{6}, - \frac{2}{3 3} + \frac{2}{3} + 3), Minimum (\frac{1}{6}, \frac{2}{3 3} - \frac{2}{3} + 3)

Schritte zur Lösung

f^{'} (x) = 12 x^{2} - 2

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{1}{6},

Absteigend

: - \frac{1}{6} < x < \frac{1}{6},

Ansteigend

: \frac{1}{6} < x < \infty

Stecke

x = - \frac{1}{6}

4 x^{3} - 2 x + 3 : - \frac{2}{3 3} + \frac{2}{3} + 3

Maximum (- \frac{1}{6}, - \frac{2}{3 3} + \frac{2}{3} + 3)

Stecke

x = \frac{1}{6}

4 x^{3} - 2 x + 3 : \frac{2}{3 3} - \frac{2}{3} + 3

Minimum (\frac{1}{6}, \frac{2}{3 3} - \frac{2}{3} + 3)

Maximum (- \frac{1}{6}, - \frac{2}{3 3} + \frac{2}{3} + 3), Minimum (\frac{1}{6}, \frac{2}{3 3} - \frac{2}{3} + 3)

e x t re m e f (x, y) = x^{2} + x y + y^{2} - 16 y + 85

e x t re m e f (x) = 4 - 5 x + x^{2}

e x t re m e f (x, y) = x^{2} + x y + y^{2} - 19 y + 120

e x t re m e f (x) = 5 - 4 x - x^{3}

e x t re m e f (x) = x^{3} + 6 x^{2} - 5, - 5 \leq x \leq 2