extreme f(x)=(x-2)(x-5)^3+4

Lösung

extrempunkte

f (x) = (x - 2) (x - 5)^{3} + 4

Minimum (\frac{11}{4}, - \frac{1163}{256}), Sattel (5, 4)

Schritte zur Lösung

f^{'} (x) = 4 x^{3} - 51 x^{2} + 210 x - 275

Intervalle finden:

Absteigend

: - \infty < x < \frac{11}{4},

Ansteigend

: \frac{11}{4} < x < 5,

Ansteigend

: 5 < x < \infty

Stecke

x = \frac{11}{4}

(x - 2) (x - 5)^{3} + 4 : - \frac{1163}{256}

Minimum (\frac{11}{4}, - \frac{1163}{256})

Stecke

x = 5

(x - 2) (x - 5)^{3} + 4 : 4

Sattel (5, 4)

Minimum (\frac{11}{4}, - \frac{1163}{256}), Sattel (5, 4)

in v erse \frac{6 x}{x + 5}

e x t re m e y = 4 x^{2} - 16 x + 11

s hi f t \frac{1}{2} cos (3 x + \frac{π}{2})

s im pl i f y (- 2.4) (7)

in f l ec t i o n \frac{3}{4} \cdot (x^{2} - 1)^{\frac{2}{3}}