extreme f(x)=(4x)/(x^2+25)

Lösung

extrempunkte

f (x) = \frac{4 x}{x ^{2} + 25}

Minimum (- 5, - \frac{2}{5}), Maximum (5, \frac{2}{5})

Schritte zur Lösung

f^{'} (x) = \frac{4 ( - x ^{2} + 25 )}{( x ^{2} + 25 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < - 5,

Ansteigend

: - 5 < x < 5,

Absteigend

: 5 < x < \infty

Stecke

x = - 5

\frac{4 x}{x ^{2} + 25} : - \frac{2}{5}

Minimum (- 5, - \frac{2}{5})

Stecke

x = 5

\frac{4 x}{x ^{2} + 25} : \frac{2}{5}

Maximum (5, \frac{2}{5})

Minimum (- 5, - \frac{2}{5}), Maximum (5, \frac{2}{5})

e x t re m e y = x^{2} - 3 x

e x t re m e f (x, y) = x^{2} + y^{4} + 2 x y

e x t re m e f (x, y) = x^{4} + y^{4} - 4 x y + 4

e x t re m e f (x) = x^{3} - 12 x - 10

e x t re m e g (x) = x^{2} - 6 x - 12