extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-6

Lösung

extrempunkte

f (x) = \frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 6

Maximum (0, - 6), Minimum (1, - \frac{94}{15})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: 0 < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = 0

\frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 6 : - 6

Maximum (0, - 6)

Stecke

x = 1

\frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 6 : - \frac{94}{15}

Minimum (1, - \frac{94}{15})

Maximum (0, - 6), Minimum (1, - \frac{94}{15})

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

e x t re m e f (x) = 16 x + \frac{9}{x}

e x t re m e f (x) = x + \frac{22}{x}

e x t re m e y = 6 x - 7 x^{\frac{6}{7}}

e x t re m e f (x, y) = ln (x^{2} + y^{2} - 64)