extreme f(x)=x^2+(12)/x

Lösung

extrempunkte

f (x) = x^{2} + \frac{12}{x}

Minimum (36, 3 \cdot 6^{\frac{2}{3}})

Schritte zur Lösung

f^{'} (x) = 2 x - \frac{12}{x ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Absteigend

: 0 < x < 36,

Ansteigend

: 36 < x < \infty

Stecke

x = 36

x^{2} + \frac{12}{x} : 3 \cdot 6^{\frac{2}{3}}

Minimum (36, 3 \cdot 6^{\frac{2}{3}})

e x t re m e f (x) = 7 cos (x), 0 \leq x \leq 2 π

e x t re m e f (x, y) = - 7 x^{2} - 3 y^{2} + 7 x - 6 y + 8

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

e x t re m e f (x, y) = x^{3} + y^{3} - 300 x - 75 y - 3

e x t re m e f (x, y) = 2 x^{3}