extreme y= x/(x^2+16)

Lösung

extrempunkte

y = \frac{x}{x ^{2} + 16}

Minimum (- 4, - \frac{1}{8}), Maximum (4, \frac{1}{8})

Schritte zur Lösung

f^{'} (x) = \frac{- x ^{2} + 16}{( x ^{2} + 16 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < - 4,

Ansteigend

: - 4 < x < 4,

Absteigend

: 4 < x < \infty

Stecke

x = - 4

\frac{x}{x ^{2} + 16} : - \frac{1}{8}

Minimum (- 4, - \frac{1}{8})

Stecke

x = 4

\frac{x}{x ^{2} + 16} : \frac{1}{8}

Maximum (4, \frac{1}{8})

Minimum (- 4, - \frac{1}{8}), Maximum (4, \frac{1}{8})

e x t re m e - x^{3} (x - 3)

e x t re m e f (x) = \frac{4}{3} x^{3} - \frac{13}{2} x^{2} - 3 x + 11

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

e x t re m e f (x) = 8 x - 16 cos (x)

e x t re m e f (x, y) = x^{2} + x y + y^{2} - 9