extreme f(x)= 1/3 x^3-3x^2+8x

Lösung

extrempunkte

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

Maximum (2, \frac{20}{3}), Minimum (4, \frac{16}{3})

Schritte zur Lösung

f^{'} (x) = x^{2} - 6 x + 8

Intervalle finden:

Ansteigend

: - \infty < x < 2,

Absteigend

: 2 < x < 4,

Ansteigend

: 4 < x < \infty

Stecke

x = 2

\frac{1}{3} x^{3} - 3 x^{2} + 8 x : \frac{20}{3}

Maximum (2, \frac{20}{3})

Stecke

x = 4

\frac{1}{3} x^{3} - 3 x^{2} + 8 x : \frac{16}{3}

Minimum (4, \frac{16}{3})

Maximum (2, \frac{20}{3}), Minimum (4, \frac{16}{3})

in t erce pt s \frac{1}{x - 6}

in v erse 2 x + 2

d o main f (x) = - 3 - \frac{36}{x - 2}

in f l ec t i o n f (x) = (x - 1)^{2} (x - 2)^{3}

in t erce pt s f (x) = x^{3} - 2 x^{2} + 2 x - 1