extreme f(x)=x(ln(x))^5

Lösung

extrempunkte

f (x) = x (ln (x))^{5}

Minimum (\frac{1}{e ^{5}}, - \frac{3125}{e ^{5}}), Sattel (1, 0)

Schritte zur Lösung

f^{'} (x) = ln^{5} (x) + 5 ln^{4} (x)

Intervalle finden:

Absteigend

: 0 < x < \frac{1}{e ^{5}},

Ansteigend

: \frac{1}{e ^{5}} < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = \frac{1}{e ^{5}}

x ln^{5} (x) : - \frac{3125}{e ^{5}}

Minimum (\frac{1}{e ^{5}}, - \frac{3125}{e ^{5}})

Stecke

x = 1

x ln^{5} (x) : 0

Sattel (1, 0)

Minimum (\frac{1}{e ^{5}}, - \frac{3125}{e ^{5}}), Sattel (1, 0)

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

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

e x t re m e x^{3} - 3 x^{2} + 2 x + 8

e x t re m e f (x) = 2 x^{2} - 4 x + y^{2} - 4 y + 6

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