extreme f(x)=x^2+(120)/x

Lösung

extrempunkte

f (x) = x^{2} + \frac{120}{x}

Minimum (360, 6 0^{\frac{2}{3}} + \frac{8 \cdot 1 5 ^{\frac{2}{3}}}{3 4})

Schritte zur Lösung

f^{'} (x) = 2 x - \frac{120}{x ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Absteigend

: 0 < x < 360,

Ansteigend

: 360 < x < \infty

Stecke

x = 360

x^{2} + \frac{120}{x} : 6 0^{\frac{2}{3}} + \frac{8 \cdot 1 5 ^{\frac{2}{3}}}{3 4}

Minimum (360, 6 0^{\frac{2}{3}} + \frac{8 \cdot 1 5 ^{\frac{2}{3}}}{3 4})

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

e x t re m e f (x) = 2 x^{5} ln (x)

e x t re m e f (x) = 48 x^{3} + 57 x^{2} - 54 x + 18

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

e x t re m e f (x) = \frac{x ^{2} + 3}{x ^{2} - 16}