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

Lösung

extrempunkte

y = \frac{x ^{2} - 2 x + 1}{x ^{2} + 3}

Maximum (- 3, \frac{4}{3}), Minimum (1, 0)

Schritte zur Lösung

f^{'} (x) = \frac{2 x ^{2} + 4 x - 6}{( x ^{2} + 3 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < - 3,

Absteigend

: - 3 < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = - 3

\frac{x ^{2} - 2 x + 1}{x ^{2} + 3} : \frac{4}{3}

Maximum (- 3, \frac{4}{3})

Stecke

x = 1

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

Minimum (1, 0)

Maximum (- 3, \frac{4}{3}), Minimum (1, 0)

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

e x t re m e f (x, y) = 3 x - 5 y - 2

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

e x t re m e x^{2} + 12 + 18 x^{- 1}

e x t re m e f (x) = x \cdot y \cdot e^{x + 2 y}