extreme f(x)=sqrt(x^2+x)-x

Lösung

extrempunkte

f (x) = x^{2} + x - x

Minimum (- 1, 1), Minimum (0, 0)

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < - 1,

Ansteigend

: 0 < x < \infty

Stecke

x = - 1

x^{2} + x - x : 1

Minimum (- 1, 1)

Stecke

x = 0

x^{2} + x - x : 0

Minimum (0, 0)

Minimum (- 1, 1), Minimum (0, 0)

e x t re m e \frac{x ^{2}}{( x - 5 ) ^{2}}

e x t re m e f (x) = 6 x - 12 cos (x), - 2 \leq x \leq 0

e x t re m e f (x) = 2 x^{3} - 33 x^{2} + 144 x + 3

e x t re m e f (x) = \frac{( 2 x ^{3} + 5 x )}{( x + 7 )}

e x t re m e f (x) = 2 x^{3} - 15 x^{2} + 6