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

Lösung

extrempunkte

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

Minimum (- 1, \frac{1}{3}), Maximum (1, 3)

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < - 1,

Ansteigend

: - 1 < x < 1,

Absteigend

: 1 < x < \infty

Stecke

x = - 1

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

Minimum (- 1, \frac{1}{3})

Stecke

x = 1

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

Maximum (1, 3)

Minimum (- 1, \frac{1}{3}), Maximum (1, 3)

e x t re m e 2 x^{5} - 15 x^{4} + 30 x^{3} - 6

e x t re m e f (x, y) = 1 + x^{2} + y^{2}

e x t re m e p (a, c) = a + 8 + c

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

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