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

Lösung

extrempunkte

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

Sattel (- 2, 0), Maximum (0, 1024), Sattel (2, 0)

Schritte zur Lösung

f^{'} (x) = - 10 x (4 - x^{2})^{4}

Intervalle finden:

Ansteigend

: - \infty < x < - 2,

Ansteigend

: - 2 < x < 0,

Absteigend

: 0 < x < 2,

Absteigend

: 2 < x < \infty

Stecke

x = - 2

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

Sattel (- 2, 0)

Stecke

x = 0

(4 - x^{2})^{5} : 1024

Maximum (0, 1024)

Stecke

x = 2

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

Sattel (2, 0)

Sattel (- 2, 0), Maximum (0, 1024), Sattel (2, 0)

e x t re m e 2 x + \frac{27}{x ^{2}}

e x t re m e f (x) = 5 x^{2} 1 + x

e x t re m e f (x) = x^{3} - \frac{3}{2} x^{2} - 18 x

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

e x t re m e f (x) = 3 x^{2} - 4 x + 5