extreme F(x)=(5x^2)/(x^2-16)

Lösung

extrempunkte

F (x) = \frac{5 x ^{2}}{x ^{2} - 16}

Maximum (0, 0)

Schritte zur Lösung

f^{'} (x) = - \frac{160 x}{( x ^{2} - 16 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < - 4,

Ansteigend

: - 4 < x < 0,

Absteigend

: 0 < x < 4,

Absteigend

: 4 < x < \infty

Stecke

x = 0

\frac{5 x ^{2}}{x ^{2} - 16} : 0

Maximum (0, 0)

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

e x t re m e 16 (- \frac{1}{x ^{2}} - \frac{2}{x ^{3}})

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

e x t re m e f (x) = 3 x^{6} - 7 x^{5}

e x t re m e f (x, y) = e^{x^{2} + 3 y^{2} + 15}