extreme f(x)= 1/3 x^3+5/2 x^2-24x+2

Lösung

extrempunkte

f (x) = \frac{1}{3} x^{3} + \frac{5}{2} x^{2} - 24 x + 2

Maximum (- 8, \frac{550}{3}), Minimum (3, - \frac{77}{2})

Schritte zur Lösung

f^{'} (x) = x^{2} + 5 x - 24

Intervalle finden:

Ansteigend

: - \infty < x < - 8,

Absteigend

: - 8 < x < 3,

Ansteigend

: 3 < x < \infty

Stecke

x = - 8

\frac{1}{3} x^{3} + \frac{5}{2} x^{2} - 24 x + 2 : \frac{550}{3}

Maximum (- 8, \frac{550}{3})

Stecke

x = 3

\frac{1}{3} x^{3} + \frac{5}{2} x^{2} - 24 x + 2 : - \frac{77}{2}

Minimum (3, - \frac{77}{2})

Maximum (- 8, \frac{550}{3}), Minimum (3, - \frac{77}{2})

e x t re m e f (x) = 4 - x

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

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

e x t re m e f (x) = x y - 2 x - y

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