extreme f(x)=((x+1)^3)/((x-1)^2)

Lösung

extrempunkte

f (x) = \frac{( x + 1 ) ^{3}}{( x - 1 ) ^{2}}

Sattel (- 1, 0), Minimum (5, \frac{27}{2})

Schritte zur Lösung

f^{'} (x) = \frac{( x + 1 ) ^{2} ( x - 5 )}{( x - 1 ) ^{3}}

Intervalle finden:

Ansteigend

: - \infty < x < - 1,

Ansteigend

: - 1 < x < 1,

Absteigend

: 1 < x < 5,

Ansteigend

: 5 < x < \infty

Stecke

x = - 1

\frac{( x + 1 ) ^{3}}{( x - 1 ) ^{2}} : 0

Sattel (- 1, 0)

Stecke

x = 5

\frac{( x + 1 ) ^{3}}{( x - 1 ) ^{2}} : \frac{27}{2}

Minimum (5, \frac{27}{2})

Sattel (- 1, 0), Minimum (5, \frac{27}{2})

e x t re m e f (x) = x^{3} - 3 x + 1, 0 \leq x \leq 3

e x t re m e f (x) = x^{3} + 2 x^{2} - x + 8

e x t re m e f (x) = \frac{1}{3} x^{3} - 4 x^{2} + 12 x - 5

e x t re m e f (x) = x 2 - x

e x t re m e f (x, y, z) = x + y 2 + x 3