extreme f(x)=(x^2-4x+4)/(x-6)

Lösung

extrempunkte

f (x) = \frac{x ^{2} - 4 x + 4}{x - 6}

Maximum (2, 0), Minimum (10, 16)

Schritte zur Lösung

f^{'} (x) = \frac{x ^{2} - 12 x + 20}{( x - 6 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < 2,

Absteigend

: 2 < x < 6,

Absteigend

: 6 < x < 10,

Ansteigend

: 10 < x < \infty

Stecke

x = 2

\frac{x ^{2} - 4 x + 4}{x - 6} : 0

Maximum (2, 0)

Stecke

x = 10

\frac{x ^{2} - 4 x + 4}{x - 6} : 16

Minimum (10, 16)

Maximum (2, 0), Minimum (10, 16)

e x t re m e f (x) = x^{3} - 588 x

e x t re m e f (x) = 23 n - 166 n^{2}

e x t re m e f (x) = \frac{- 2}{3} x^{3} + 9 x^{2} - 28 x + 1

e x t re m e y = \frac{1}{x ^{2} - 1}

e x t re m e f (x, y) = (x^{2} - 25)^{2} + (y^{2} - 9)^{2}