extreme f(x)=8+6x-7x^2-4x^3

Lösung

extrempunkte

f (x) = 8 + 6 x - 7 x^{2} - 4 x^{3}

Minimum (- \frac{3}{2}, - \frac{13}{4}), Maximum (\frac{1}{3}, \frac{245}{27})

Schritte zur Lösung

f^{'} (x) = - 12 x^{2} - 14 x + 6

Intervalle finden:

Absteigend

: - \infty < x < - \frac{3}{2},

Ansteigend

: - \frac{3}{2} < x < \frac{1}{3},

Absteigend

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

Stecke

x = - \frac{3}{2}

8 + 6 x - 7 x^{2} - 4 x^{3} : - \frac{13}{4}

Minimum (- \frac{3}{2}, - \frac{13}{4})

Stecke

x = \frac{1}{3}

8 + 6 x - 7 x^{2} - 4 x^{3} : \frac{245}{27}

Maximum (\frac{1}{3}, \frac{245}{27})

Minimum (- \frac{3}{2}, - \frac{13}{4}), Maximum (\frac{1}{3}, \frac{245}{27})

e x t re m e f (x) = \frac{20}{3 ( 2 + x ) ^{3}}

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

e x t re m e 2 (x - 1) (x^{2} + 1) (3 x^{2} - 2 x + 1)

e x t re m e x^{3} + 30 x + 128

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