extreme f(x)=-2/3 x^3+2x^2+6x-50

Lösung

extrempunkte

f (x) = - \frac{2}{3} x^{3} + 2 x^{2} + 6 x - 50

Minimum (- 1, - \frac{160}{3}), Maximum (3, - 32)

Schritte zur Lösung

f^{'} (x) = - 2 x^{2} + 4 x + 6

Intervalle finden:

Absteigend

: - \infty < x < - 1,

Ansteigend

: - 1 < x < 3,

Absteigend

: 3 < x < \infty

Stecke

x = - 1

- \frac{2}{3} x^{3} + 2 x^{2} + 6 x - 50 : - \frac{160}{3}

Minimum (- 1, - \frac{160}{3})

Stecke

x = 3

- \frac{2}{3} x^{3} + 2 x^{2} + 6 x - 50 : - 32

Maximum (3, - 32)

Minimum (- 1, - \frac{160}{3}), Maximum (3, - 32)

e x t re m e f (x) = x^{5} - 20 x^{4} + 8, 0 \leq x \leq 17

e x t re m e P (q, r) = 4 - q + r

\frac{d}{d x} (I n 1 - 2 x)

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

e x t re m e \frac{e ^{3.4 x}}{x ^{3}}