extreme (3-x)(x-5)^2

Lösung

extrempunkte

(3 - x) (x - 5)^{2}

Minimum (\frac{11}{3}, - \frac{32}{27}), Maximum (5, 0)

Schritte zur Lösung

f^{'} (x) = - (x - 5) (3 x - 11)

Intervalle finden:

Absteigend

: - \infty < x < \frac{11}{3},

Ansteigend

: \frac{11}{3} < x < 5,

Absteigend

: 5 < x < \infty

Stecke

x = \frac{11}{3}

(3 - x) (x - 5)^{2} : - \frac{32}{27}

Minimum (\frac{11}{3}, - \frac{32}{27})

Stecke

x = 5

(3 - x) (x - 5)^{2} : 0

Maximum (5, 0)

Minimum (\frac{11}{3}, - \frac{32}{27}), Maximum (5, 0)

e x t re m e f (x, y) = \frac{1}{x} + \frac{1}{y} - x y + 3

e x t re m e f (x) = x^{3} + y^{3} + 9 x^{2} - 9 y^{2} - 2

e x t re m e f (x) = \frac{( x )}{( 6 x + 2 )}, 5 \leq x \leq 8

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

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