extreme f(x)= x/(x^2-x+16)

Lösung

extrempunkte

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

Minimum (- 4, - \frac{1}{9}), Maximum (4, \frac{1}{7})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < - 4,

Ansteigend

: - 4 < x < 4,

Absteigend

: 4 < x < \infty

Stecke

x = - 4

\frac{x}{x ^{2} - x + 16} : - \frac{1}{9}

Minimum (- 4, - \frac{1}{9})

Stecke

x = 4

\frac{x}{x ^{2} - x + 16} : \frac{1}{7}

Maximum (4, \frac{1}{7})

Minimum (- 4, - \frac{1}{9}), Maximum (4, \frac{1}{7})

e x t re m e f (x) = \frac{x ^{3}}{3} - 5 x^{2} + 16 x - 100

e x t re m e f (x) = x^{3} + 10 x^{2} + 8

e x t re m e f (x, y) = - 6 x^{2} + 5 x y - y^{2} + x + y

e x t re m e f (x, y) = 3 x^{2} y - y^{2} + 5 x y^{2}

e x t re m e f (x) = 2 x + cos (x)