extreme y=x^{2/5}(x+3)

Lösung

extrempunkte

y = x^{\frac{2}{5}} (x + 3)

Maximum - \frac{6}{7}, \frac{15 ( - \frac{6}{7} ) ^{\frac{2}{5}}}{7}, Minimum (0, 0)

Schritte zur Lösung

f^{'} (x) = \frac{7 x + 6}{5 x ^{\frac{3}{5}}}

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{6}{7},

Absteigend

: - \frac{6}{7} < x < 0,

Ansteigend

: 0 < x < \infty

Stecke

x = - \frac{6}{7}

x^{\frac{2}{5}} (x + 3) : \frac{15 ( - \frac{6}{7} ) ^{\frac{2}{5}}}{7}

Maximum - \frac{6}{7}, \frac{15 ( - \frac{6}{7} ) ^{\frac{2}{5}}}{7}

Stecke

x = 0

x^{\frac{2}{5}} (x + 3) : 0

Minimum (0, 0)

Maximum - \frac{6}{7}, \frac{15 ( - \frac{6}{7} ) ^{\frac{2}{5}}}{7}, Minimum (0, 0)

e x t re m e x + 2 cos (x)

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

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

e x t re m e f (x, y) = - x^{3} + 4 x \cdot y - 2 y^{2} + 1

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