extreme f(1)=x^3-6x^2+9x+2

Lösung

extrempunkte

f (1) = x^{3} - 6 x^{2} + 9 x + 2

Maximum (1, 6), Minimum (3, 2)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} - 12 x + 9

Intervalle finden:

Ansteigend

: - \infty < x < 1,

Absteigend

: 1 < x < 3,

Ansteigend

: 3 < x < \infty

Stecke

x = 1

x^{3} - 6 x^{2} + 9 x + 2 : 6

Maximum (1, 6)

Stecke

x = 3

x^{3} - 6 x^{2} + 9 x + 2 : 2

Minimum (3, 2)

Maximum (1, 6), Minimum (3, 2)

e x t re m e f (x) = \frac{2}{3} x^{3} + \frac{11}{2} x^{2} - 21 x + 5

e x t re m e f (x) = (x - 3) e^{- 5 x}

e x t re m e f (x) = - 2^{3} - 3 \cdot (- 2^{2}) + 3 \cdot (- 2) + 1

e x t re m e g (x) = e^{- x^{4}}, - 3 \leq x \leq 1

e x t re m e f (x, y) = x^{2} + y^{2} - 4 x + 3 y