extreme f(x)=(t^2)/(1+t^3)

Lösung

extrempunkte

f (x) = \frac{t ^{2}}{1 + t ^{3}}

Minimum (0, 0), Maximum (32, \frac{2 ^{\frac{2}{3}}}{3})

Schritte zur Lösung

f^{'} (t) = \frac{2 t - t ^{4}}{( 1 + t ^{3} ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < t < - 1,

Absteigend

: - 1 < t < 0,

Ansteigend

: 0 < t < 32,

Absteigend

: 32 < t < \infty

Stecke

t = 0

\frac{t ^{2}}{1 + t ^{3}} : 0

Minimum (0, 0)

Stecke

t = 32

\frac{t ^{2}}{1 + t ^{3}} : \frac{2 ^{\frac{2}{3}}}{3}

Maximum (32, \frac{2 ^{\frac{2}{3}}}{3})

Minimum (0, 0), Maximum (32, \frac{2 ^{\frac{2}{3}}}{3})

e x t re m e - x^{3} + 6 x^{2} + x - 1

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

e x t re m e - x^{3} + 6 x^{2} + x + 1

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

e x t re m e w (x, y) = x + x y