extreme f(x)=6x^3-12x^2+5x-2

Lösung

extrempunkte

f (x) = 6 x^{3} - 12 x^{2} + 5 x - 2

Maximum (\frac{4 - 6}{6}, \frac{3 6 - 2}{9} - 2), Minimum (\frac{4 + 6}{6}, \frac{- 2 - 3 6}{9} - 2)

Schritte zur Lösung

f^{'} (x) = 18 x^{2} - 24 x + 5

Intervalle finden:

Ansteigend

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

Absteigend

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

Ansteigend

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

Stecke

x = \frac{4 - 6}{6}

6 x^{3} - 12 x^{2} + 5 x - 2 : \frac{3 6 - 2}{9} - 2

Maximum (\frac{4 - 6}{6}, \frac{3 6 - 2}{9} - 2)

Stecke

x = \frac{4 + 6}{6}

6 x^{3} - 12 x^{2} + 5 x - 2 : \frac{- 2 - 3 6}{9} - 2

Minimum (\frac{4 + 6}{6}, \frac{- 2 - 3 6}{9} - 2)

Maximum (\frac{4 - 6}{6}, \frac{3 6 - 2}{9} - 2), Minimum (\frac{4 + 6}{6}, \frac{- 2 - 3 6}{9} - 2)

in f l ec t i o n x^{3} - 12 x^{2} - 27 x + 8

cr i t i c a l f (x) = \frac{e ^{x}}{3 + e ^{x}}

p er i o d i c i t y y = sin (x + \frac{π}{2})

r an g e - x^{2} + 8 x - 7

in v erse f (x) = \frac{( x - 4 )}{( x - 2 )}