extreme f(x)=(2+4x^2)/(1-3x)

Lösung

extrempunkte

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

Minimum (- \frac{- 2 + 22}{6}, \frac{44 2 - 8 11}{9 11}), Maximum (\frac{2 + 22}{6}, - \frac{44 2 + 8 11}{9 11})

Schritte zur Lösung

f^{'} (x) = \frac{- 12 x ^{2} + 8 x + 6}{( 1 - 3 x ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < \frac{2 - 22}{6},

Ansteigend

: \frac{2 - 22}{6} < x < \frac{1}{3},

Ansteigend

: \frac{1}{3} < x < \frac{2 + 22}{6},

Absteigend

: \frac{2 + 22}{6} < x < \infty

Stecke

x = - \frac{- 2 + 22}{6}

\frac{2 + 4 x ^{2}}{1 - 3 x} : \frac{44 2 - 8 11}{9 11}

Minimum (- \frac{- 2 + 22}{6}, \frac{44 2 - 8 11}{9 11})

Stecke

x = \frac{2 + 22}{6}

\frac{2 + 4 x ^{2}}{1 - 3 x} : - \frac{44 2 + 8 11}{9 11}

Maximum (\frac{2 + 22}{6}, - \frac{44 2 + 8 11}{9 11})

Minimum (- \frac{- 2 + 22}{6}, \frac{44 2 - 8 11}{9 11}), Maximum (\frac{2 + 22}{6}, - \frac{44 2 + 8 11}{9 11})

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}

e x t re m e f (x) = \frac{350}{( 1 + 34 e ^{- t} )}