extreme f(x)=x^2-4x+16

Lösung

extrempunkte

f (x) = x^{2} - 4 x + 16

Minimum (2, 12)

Schritte zur Lösung

f^{'} (x) = 2 x - 4

Intervalle finden:

Absteigend

: - \infty < x < 2,

Ansteigend

: 2 < x < \infty

Stecke

x = 2

x^{2} - 4 x + 16 : 12

Minimum (2, 12)

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

e x t re m e f (x) = (x^{2} + 1)^{\frac{2}{3}}

e x t re m e f (x) = 2 x^{4} - 7 x^{3} + 4 x^{2} + 2 x + 3

e x t re m e 4 x^{3} - 3 x^{2} - 36 x + 17

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