extreme f(x)= x/(x^2-x+9)

Lösung

extrempunkte

f (x) = \frac{x}{x ^{2} - x + 9}

Minimum (- 3, - \frac{1}{7}), Maximum (3, \frac{1}{5})

Schritte zur Lösung

f^{'} (x) = \frac{- x ^{2} + 9}{( x ^{2} - x + 9 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < - 3,

Ansteigend

: - 3 < x < 3,

Absteigend

: 3 < x < \infty

Stecke

x = - 3

\frac{x}{x ^{2} - x + 9} : - \frac{1}{7}

Minimum (- 3, - \frac{1}{7})

Stecke

x = 3

\frac{x}{x ^{2} - x + 9} : \frac{1}{5}

Maximum (3, \frac{1}{5})

Minimum (- 3, - \frac{1}{7}), Maximum (3, \frac{1}{5})

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

e x t re m e 9 - 2 x + 4 y - x^{2} - 4 y^{2}

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

e x t re m e f (x) = - 5 cos (x - 3 π)

e x t re m e f (x) = - 3 x^{4} - 7 x^{3} + 23