extreme f(x)=x*(ln(x))^2

Lösung

extrempunkte

f (x) = x \cdot (ln (x))^{2}

Maximum (\frac{1}{e ^{2}}, \frac{4}{e ^{2}}), Minimum (1, 0)

Schritte zur Lösung

f^{'} (x) = ln^{2} (x) + 2 ln (x)

Intervalle finden:

Ansteigend

: 0 < x < \frac{1}{e ^{2}},

Absteigend

: \frac{1}{e ^{2}} < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = \frac{1}{e ^{2}}

x \cdot ln^{2} (x) : \frac{4}{e ^{2}}

Maximum (\frac{1}{e ^{2}}, \frac{4}{e ^{2}})

Stecke

x = 1

x \cdot ln^{2} (x) : 0

Minimum (1, 0)

Maximum (\frac{1}{e ^{2}}, \frac{4}{e ^{2}}), Minimum (1, 0)

e x t re m e f (x) = x [36 - 2 (x)]^{2}

e x t re m e f (x) = x + 12 - 8

e x t re m e f (x, y) = 4 x y + 120 y - 20 y^{2} - \frac{1}{10} x^{2} y - 90

e x t re m e f (x, y) = 2 x^{2} + 4 y^{2} - 2 x y - 10 x - 2 y + 2

e x t re m e f (x) = \frac{7 x}{x ^{2} + 9}, 0 \leq x \leq 9