extreme f(x)=x^2-8x+15

Lösung

extrempunkte

f (x) = x^{2} - 8 x + 15

Minimum (4, - 1)

Schritte zur Lösung

f^{'} (x) = 2 x - 8

Intervalle finden:

Absteigend

: - \infty < x < 4,

Ansteigend

: 4 < x < \infty

Stecke

x = 4

x^{2} - 8 x + 15 : - 1

Minimum (4, - 1)

e x t re m e f (x) = 3 x^{4} - 4 x^{3} - 36 x^{2} + 5

e x t re m e f (x) = x + lo g_{10} (1 - x)

e x t re m e f (x) = - x^{4} + 16 x^{3} - 16 x + 11

e x t re m e f (x) = x^{3} + x^{2} - 5 x

e x t re m e f (x, y) = (x + 2 y - 7)^{2} + (2 x + y - 5)^{2}