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

Lösung

extrempunkte

f (x) = \frac{x ^{3}}{3} + x^{2} - 8 x + 2

Maximum (- 4, \frac{86}{3}), Minimum (2, - \frac{22}{3})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 4,

Absteigend

: - 4 < x < 2,

Ansteigend

: 2 < x < \infty

Stecke

x = - 4

\frac{x ^{3}}{3} + x^{2} - 8 x + 2 : \frac{86}{3}

Maximum (- 4, \frac{86}{3})

Stecke

x = 2

\frac{x ^{3}}{3} + x^{2} - 8 x + 2 : - \frac{22}{3}

Minimum (2, - \frac{22}{3})

Maximum (- 4, \frac{86}{3}), Minimum (2, - \frac{22}{3})

e x t re m e E (a, b) = - [a - b - (a - (b - a - (a - (- b - a))))] - b

e x t re m e ∣ x - 5 ∣ + 1

e x t re m e f (x) = x (10 + x), - 1 \leq x \leq 3

e x t re m e y = x (10 - 2 x) (8 - x)

e x t re m e f (x) = \frac{3 x ^{2} - 16 x + 16}{( x ^{2} - 2 x ) ^{2}}