extreme f(x)=(x^3-27)/(x^2-9)

Lösung

extrempunkte

f (x) = \frac{x ^{3} - 27}{x ^{2} - 9}

Maximum (- 6, - 9), Minimum (0, 3)

Schritte zur Lösung

f^{'} (x) = \frac{x ( x + 6 )}{( x + 3 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < - 6,

Absteigend

: - 6 < x < - 3,

Absteigend

: - 3 < x < 0,

Ansteigend

: 0 < x < 3,

Ansteigend

: 3 < x < \infty

Stecke

x = - 6

\frac{x ^{3} - 27}{x ^{2} - 9} : - 9

Maximum (- 6, - 9)

Stecke

x = 0

\frac{x ^{3} - 27}{x ^{2} - 9} : 3

Minimum (0, 3)

Maximum (- 6, - 9), Minimum (0, 3)

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

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

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

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

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