extreme f(x)=(x^2-2x+1)/(x-4)

Lösung

extrempunkte

f (x) = \frac{x ^{2} - 2 x + 1}{x - 4}

Maximum (1, 0), Minimum (7, 12)

Schritte zur Lösung

f^{'} (x) = \frac{x ^{2} - 8 x + 7}{( x - 4 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < 1,

Absteigend

: 1 < x < 4,

Absteigend

: 4 < x < 7,

Ansteigend

: 7 < x < \infty

Stecke

x = 1

\frac{x ^{2} - 2 x + 1}{x - 4} : 0

Maximum (1, 0)

Stecke

x = 7

\frac{x ^{2} - 2 x + 1}{x - 4} : 12

Minimum (7, 12)

Maximum (1, 0), Minimum (7, 12)

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

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

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

e x t re m e f (x) = (x - 3) (x - 7)^{3} + 12, (1, 4)

e x t re m e f (x, y) = 2 x^{3} + y^{2} - 6 x y - 23