extreme f(x)=(x^2-x-2)/(x^2-6x+9)

Lösung

extrempunkte

f (x) = \frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9}

Minimum (\frac{7}{5}, - \frac{9}{16})

Schritte zur Lösung

f^{'} (x) = - \frac{5 x - 7}{( x - 3 ) ^{3}}

Intervalle finden:

Absteigend

: - \infty < x < \frac{7}{5},

Ansteigend

: \frac{7}{5} < x < 3,

Absteigend

: 3 < x < \infty

Stecke

x = \frac{7}{5}

\frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9} : - \frac{9}{16}

Minimum (\frac{7}{5}, - \frac{9}{16})

e x t re m e f (x, y) = x y + \frac{8}{x} + \frac{8}{y}

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

e x t re m e f (x, y) = - 5 x^{2} + 4 x y - y^{2} + 16 x + 10

e x t re m e f (x, y) = \frac{y}{x}

e x t re m e f (x, y) = \frac{x ^{2} - 5 y ^{2}}{x + 4}