extreme f(x)=x^3-(1/2)x^2-4x+12

Lösung

extrempunkte

f (x) = x^{3} - (\frac{1}{2}) x^{2} - 4 x + 12

Maximum (- 1, \frac{29}{2}), Minimum (\frac{4}{3}, \frac{220}{27})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 1,

Absteigend

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

Ansteigend

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

Stecke

x = - 1

x^{3} - \frac{1}{2} x^{2} - 4 x + 12 : \frac{29}{2}

Maximum (- 1, \frac{29}{2})

Stecke

x = \frac{4}{3}

x^{3} - \frac{1}{2} x^{2} - 4 x + 12 : \frac{220}{27}

Minimum (\frac{4}{3}, \frac{220}{27})

Maximum (- 1, \frac{29}{2}), Minimum (\frac{4}{3}, \frac{220}{27})

e x t re m e f (x) = 2 x^{3} + 3 x^{2} - 36 x + 1 [- 6.3]

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

e x t re m e f (x) = 4 x^{3} + 6 x^{2} - 24 x

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

e x t re m e f (x) = - 4 x^{6} ln (3 x)