extreme f(x)=(2x^{5/2})/5-(2x^{3/2})/3-1

Lösung

extrempunkte

f (x) = \frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 1

Maximum (0, - 1), Minimum (1, - \frac{19}{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} - 1 : - 1

Maximum (0, - 1)

Stecke

x = 1

\frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 1 : - \frac{19}{15}

Minimum (1, - \frac{19}{15})

Maximum (0, - 1), Minimum (1, - \frac{19}{15})

e x t re m e 5 x^{3} - 5 x^{2} - 5 x + 9

e x t re m e \frac{x ^{3}}{x ^{2} - 49}

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

e x t re m e 4 + x^{3} + y^{3} - 3 x y

e x t re m e p (x) = 3 x^{3} - 13 x^{2} + a x - 8