extreme f(x)=(x^6}{10}-\frac{2x^5)/3

Lösung

extrempunkte

f (x) = \frac{x ^{6}}{10} - \frac{2 x ^{5}}{3}

Sattel (0, 0), Minimum (\frac{50}{9}, - \frac{312500000}{531441})

Schritte zur Lösung

f^{'} (x) = \frac{3 x ^{5}}{5} - \frac{10 x ^{4}}{3}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Absteigend

: 0 < x < \frac{50}{9},

Ansteigend

: \frac{50}{9} < x < \infty

Stecke

x = 0

\frac{x ^{6}}{10} - \frac{2 x ^{5}}{3} : 0

Sattel (0, 0)

Stecke

x = \frac{50}{9}

\frac{x ^{6}}{10} - \frac{2 x ^{5}}{3} : - \frac{312500000}{531441}

Minimum (\frac{50}{9}, - \frac{312500000}{531441})

Sattel (0, 0), Minimum (\frac{50}{9}, - \frac{312500000}{531441})

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

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

e x t re m e f (x, y) = x^{2} + x y + y^{2} - 31 y + 320

e x t re m e f (x, y) = 5 x + 3 y

e x t re m e f (x) = x^{3} + 8 x^{2} + 16 x + 8, - 7 \leq x \leq 0