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

Lösung

extrempunkte

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

Minimum (- 2 (1 + 2), - \frac{2 - 1}{2}), Maximum (2 (2 - 1), \frac{2 + 1}{2})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < - 2 (1 + 2),

Ansteigend

: - 2 (1 + 2) < x < 2 (2 - 1),

Absteigend

: 2 (2 - 1) < x < \infty

Stecke

x = - 2 (1 + 2)

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

Minimum (- 2 (1 + 2), - \frac{2 - 1}{2})

Stecke

x = 2 (2 - 1)

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

Maximum (2 (2 - 1), \frac{2 + 1}{2})

Minimum (- 2 (1 + 2), - \frac{2 - 1}{2}), Maximum (2 (2 - 1), \frac{2 + 1}{2})

e x t re m e f (x) = (6 x + 1) e^{8 x - x^{2}}

e x t re m e f (x) = x^{4} - 3 x^{2}, - 2 \leq x \leq 0

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

e x t re m e y = 15 x e^{- 0.05 x} - 30

e x t re m e f (x) = \frac{- 4 x}{x ^{2} + 6}, (0, 6)