extreme y= x/(x^2+64)

Lösung

extrempunkte

y = \frac{x}{x ^{2} + 64}

Minimum (- 8, - \frac{1}{16}), Maximum (8, \frac{1}{16})

Schritte zur Lösung

f^{'} (x) = \frac{- x ^{2} + 64}{( x ^{2} + 64 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < - 8,

Ansteigend

: - 8 < x < 8,

Absteigend

: 8 < x < \infty

Stecke

x = - 8

\frac{x}{x ^{2} + 64} : - \frac{1}{16}

Minimum (- 8, - \frac{1}{16})

Stecke

x = 8

\frac{x}{x ^{2} + 64} : \frac{1}{16}

Maximum (8, \frac{1}{16})

Minimum (- 8, - \frac{1}{16}), Maximum (8, \frac{1}{16})

e x t re m e f (x) = - x^{2} + 1, - 1 \leq x \leq 0

e x t re m e f (x) = x^{5} - x^{3} + 8, - 1 \leq x \leq 1

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

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

e x t re m e f (x) = 12 x + 6