extreme f(x)=-(15)/(x^2-19x+16)

Lösung

extrempunkte

f (x) = - \frac{15}{x ^{2} - 19 x + 16}

Minimum (\frac{19}{2}, \frac{20}{99})

Schritte zur Lösung

f^{'} (x) = \frac{15 ( 2 x - 19 )}{( x ^{2} - 19 x + 16 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < \frac{19 - 3 33}{2},

Absteigend

: \frac{19 - 3 33}{2} < x < \frac{19}{2},

Ansteigend

: \frac{19}{2} < x < \frac{19 + 3 33}{2},

Ansteigend

: \frac{19 + 3 33}{2} < x < \infty

Stecke

x = \frac{19}{2}

- \frac{15}{x ^{2} - 19 x + 16} : \frac{20}{99}

Minimum (\frac{19}{2}, \frac{20}{99})

e x t re m e f (x) = x + \frac{9}{x}, 0.2 \leq x \leq 12

e x t re m e f (x) = x^{2} - 6 x - 1

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

e x t re m e f (x, y) = \frac{- 2 x}{y} + e^{x} - 13

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