extreme f(x)=2x^4+6x^3-12x^2+8

Lösung

extrempunkte

f (x) = 2 x^{4} + 6 x^{3} - 12 x^{2} + 8

Minimum (\frac{- 9 - 273}{8}, - \frac{9 ( 1619 + 91 273 )}{256} + 8), Maximum (0, 8), Minimum (\frac{- 9 + 273}{8}, \frac{9 ( 91 273 - 1619 )}{256} + 8)

Schritte zur Lösung

f^{'} (x) = 8 x^{3} + 18 x^{2} - 24 x

Intervalle finden:

Absteigend

: - \infty < x < \frac{- 9 - 273}{8},

Ansteigend

: \frac{- 9 - 273}{8} < x < 0,

Absteigend

: 0 < x < \frac{273 - 9}{8},

Ansteigend

: \frac{273 - 9}{8} < x < \infty

Stecke

x = \frac{- 9 - 273}{8}

2 x^{4} + 6 x^{3} - 12 x^{2} + 8 : - \frac{9 ( 1619 + 91 273 )}{256} + 8

Minimum (\frac{- 9 - 273}{8}, - \frac{9 ( 1619 + 91 273 )}{256} + 8)

Stecke

x = 0

2 x^{4} + 6 x^{3} - 12 x^{2} + 8 : 8

Maximum (0, 8)

Stecke

x = \frac{- 9 + 273}{8}

2 x^{4} + 6 x^{3} - 12 x^{2} + 8 : \frac{9 ( 91 273 - 1619 )}{256} + 8

Minimum (\frac{- 9 + 273}{8}, \frac{9 ( 91 273 - 1619 )}{256} + 8)

Minimum (\frac{- 9 - 273}{8}, - \frac{9 ( 1619 + 91 273 )}{256} + 8), Maximum (0, 8), Minimum (\frac{- 9 + 273}{8}, \frac{9 ( 91 273 - 1619 )}{256} + 8)

m o n o t o n e f (x) = - x + 3

s l o p e in t erce pt 5 x - 15 y = 17

cr i t i c a l f (x) = (8 - 4 x) e^{x}

d o main sin (arccos (x))

l in e (- 8, 5), (17, 8)