extreme f(x)=(-1)/(5x^2+25x+30)

Lösung

extrempunkte

f (x) = \frac{- 1}{5 x ^{2} + 25 x + 30}

Minimum (- \frac{5}{2}, \frac{4}{5})

Schritte zur Lösung

f^{'} (x) = \frac{2 x + 5}{5 ( x + 2 ) ^{2} ( x + 3 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < - 3,

Absteigend

: - 3 < x < - \frac{5}{2},

Ansteigend

: - \frac{5}{2} < x < - 2,

Ansteigend

: - 2 < x < \infty

Stecke

x = - \frac{5}{2}

\frac{- 1}{5 x ^{2} + 25 x + 30} : \frac{4}{5}

Minimum (- \frac{5}{2}, \frac{4}{5})

e x t re m e f (x) = \frac{5 x}{x ^{2} - 1}

e x t re m e f (x) = - \frac{x ^{3}}{3} + 2 x^{2} + 5 x + 1

e x t re m e \frac{7 x + 5}{x + 9} + \frac{8 x - 10}{x + 9}

e x t re m e x - 8 3 x

e x t re m e f (x) = x^{2} + x y + y^{2} - 22 y + 161