extreme f(x)= 8/3 x^3+32x^2+120x+9

Lösung

extrempunkte

f (x) = \frac{8}{3} x^{3} + 32 x^{2} + 120 x + 9

Maximum (- 5, - \frac{373}{3}), Minimum (- 3, - 135)

Schritte zur Lösung

f^{'} (x) = 8 x^{2} + 64 x + 120

Intervalle finden:

Ansteigend

: - \infty < x < - 5,

Absteigend

: - 5 < x < - 3,

Ansteigend

: - 3 < x < \infty

Stecke

x = - 5

\frac{8}{3} x^{3} + 32 x^{2} + 120 x + 9 : - \frac{373}{3}

Maximum (- 5, - \frac{373}{3})

Stecke

x = - 3

\frac{8}{3} x^{3} + 32 x^{2} + 120 x + 9 : - 135

Minimum (- 3, - 135)

Maximum (- 5, - \frac{373}{3}), Minimum (- 3, - 135)

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e x t re m e f (x) = 2 θ - 3 sin (θ)

e x t re m e f (x, y) = 2 x ln (y^{2} - 4)

e x t re m e f (x, y) = 6 x - 3 y + 4

e x t re m e f (x) = 3 θ - 6 sin (θ)