extreme f(x)=9x^3+21x^2+8x-3

Lösung

extrempunkte

f (x) = 9 x^{3} + 21 x^{2} + 8 x - 3

Maximum (- \frac{4}{3}, \frac{7}{3}), Minimum (- \frac{2}{9}, - \frac{311}{81})

Schritte zur Lösung

f^{'} (x) = 27 x^{2} + 42 x + 8

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{4}{3},

Absteigend

: - \frac{4}{3} < x < - \frac{2}{9},

Ansteigend

: - \frac{2}{9} < x < \infty

Stecke

x = - \frac{4}{3}

9 x^{3} + 21 x^{2} + 8 x - 3 : \frac{7}{3}

Maximum (- \frac{4}{3}, \frac{7}{3})

Stecke

x = - \frac{2}{9}

9 x^{3} + 21 x^{2} + 8 x - 3 : - \frac{311}{81}

Minimum (- \frac{2}{9}, - \frac{311}{81})

Maximum (- \frac{4}{3}, \frac{7}{3}), Minimum (- \frac{2}{9}, - \frac{311}{81})

e x t re m e f (x) = x^{2} - 2 x + 5 [- 1.4]

e x t re m e 2 x^{4} - 196 x^{2} - 3

e x t re m e f (x) = - x^{2} - y^{2} + 8 x + 8 y

e x t re m e f (x, y) = x^{3} + 8 x^{2} y^{2} - 8 y^{3} - x + y

e x t re m e f (t) = 4 t^{3} - 36 t^{2} + 10, t \geq 0