extreme f(x)=((2x^2+4x))/((2+x^2))

Lösung

extrempunkte

f (x) = \frac{( 2 x ^{2} + 4 x )}{( 2 + x ^{2} )}

Minimum (1 - 3, 1 - 3), Maximum (1 + 3, 1 + 3)

Schritte zur Lösung

f^{'} (x) = \frac{- 4 x ^{2} + 8 x + 8}{( 2 + x ^{2} ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < - 3 + 1,

Ansteigend

: - 3 + 1 < x < 1 + 3,

Absteigend

: 1 + 3 < x < \infty

Stecke

x = 1 - 3

\frac{2 x ^{2} + 4 x}{2 + x ^{2}} : 1 - 3

Minimum (1 - 3, 1 - 3)

Stecke

x = 1 + 3

\frac{2 x ^{2} + 4 x}{2 + x ^{2}} : 1 + 3

Maximum (1 + 3, 1 + 3)

Minimum (1 - 3, 1 - 3), Maximum (1 + 3, 1 + 3)

e x t re m e f (x) = x^{3} + 8 x + 12

e x t re m e 7 - 6 x - x^{3}

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

e x t re m e f (x) = x^{2} - 14 x + 4

e x t re m e f (x) = x^{2} - 14 x + 6