extreme f(x)=4x^3+9x^2-12x+7

Lösung

extrempunkte

f (x) = 4 x^{3} + 9 x^{2} - 12 x + 7

Maximum (- 2, 35), Minimum (\frac{1}{2}, \frac{15}{4})

Schritte zur Lösung

f^{'} (x) = 12 x^{2} + 18 x - 12

Intervalle finden:

Ansteigend

: - \infty < x < - 2,

Absteigend

: - 2 < x < \frac{1}{2},

Ansteigend

: \frac{1}{2} < x < \infty

Stecke

x = - 2

4 x^{3} + 9 x^{2} - 12 x + 7 : 35

Maximum (- 2, 35)

Stecke

x = \frac{1}{2}

4 x^{3} + 9 x^{2} - 12 x + 7 : \frac{15}{4}

Minimum (\frac{1}{2}, \frac{15}{4})

Maximum (- 2, 35), Minimum (\frac{1}{2}, \frac{15}{4})

e x t re m e h (v, t) = v t - 16 t^{2}

e x t re m e f (x) = \frac{1}{2} x^{3} - 3 x y + 3 y^{2} - x

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

e x t re m e f (x, y) = 12 - x^{2} - y^{2}

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