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

Lösung

extrempunkte

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

Maximum (- 3, - 8), Minimum (1, 0)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 3,

Absteigend

: - 3 < x < - 1,

Absteigend

: - 1 < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = - 3

\frac{x ^{2} - 2 x + 1}{x + 1} : - 8

Maximum (- 3, - 8)

Stecke

x = 1

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

Minimum (1, 0)

Maximum (- 3, - 8), Minimum (1, 0)

e x t re m e f (x) = - x^{3} + 6 x^{2} - 9 x - 1

e x t re m e f (x) = (- \frac{1}{100} (x) (x - 300))

e x t re m e f (x) = 2 x + 6, - 4 \leq x \leq 2

e x t re m e f (x) = (x - 9)^{3}

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