extreme y=(t-3t^2)^{1/3}

Lösung

extrempunkte

y = (t - 3 t^{2})^{\frac{1}{3}}

Maximum (\frac{1}{6}, (\frac{1}{12})^{\frac{1}{3}})

Schritte zur Lösung

f^{'} (t) = \frac{1 - 6 t}{3 ( t - 3 t ^{2} ) ^{\frac{2}{3}}}

Intervalle finden:

Ansteigend

: - \infty < t < 0,

Ansteigend

: 0 < t < \frac{1}{6},

Absteigend

: \frac{1}{6} < t < \frac{1}{3},

Absteigend

: \frac{1}{3} < t < \infty

Stecke

t = \frac{1}{6}

(t - 3 t^{2})^{\frac{1}{3}} : (\frac{1}{12})^{\frac{1}{3}}

Maximum (\frac{1}{6}, (\frac{1}{12})^{\frac{1}{3}})

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

e x t re m e f (1) = x^{3} - 6 x^{2} + 9 x + 2

e x t re m e f (x) = \frac{2}{3} x^{3} + \frac{11}{2} x^{2} - 21 x + 5

e x t re m e f (x) = (x - 3) e^{- 5 x}

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