extreme f(x)=(t-4t^2)^{1/3}

Lösung

extrempunkte

f (x) = (t - 4 t^{2})^{\frac{1}{3}}

Maximum (\frac{1}{8}, \frac{1}{2 \cdot 2 ^{\frac{1}{3}}})

Schritte zur Lösung

f^{'} (t) = \frac{1 - 8 t}{3 ( t - 4 t ^{2} ) ^{\frac{2}{3}}}

Intervalle finden:

Ansteigend

: - \infty < t < 0,

Ansteigend

: 0 < t < \frac{1}{8},

Absteigend

: \frac{1}{8} < t < \frac{1}{4},

Absteigend

: \frac{1}{4} < t < \infty

Stecke

t = \frac{1}{8}

(t - 4 t^{2})^{\frac{1}{3}} : \frac{1}{2 \cdot 2 ^{\frac{1}{3}}}

Maximum (\frac{1}{8}, \frac{1}{2 \cdot 2 ^{\frac{1}{3}}})

e x t re m e f (x, y) = 1 - (x^{2} + y^{2})

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

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

e x t re m e f (x) = 4 x^{5} - 7 x^{2} + 8

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