extreme y=(6x)/(x^2+4)

Lösung

extrempunkte

y = \frac{6 x}{x ^{2} + 4}

Minimum (- 2, - \frac{3}{2}), Maximum (2, \frac{3}{2})

Schritte zur Lösung

f^{'} (x) = \frac{6 ( - x ^{2} + 4 )}{( x ^{2} + 4 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < - 2,

Ansteigend

: - 2 < x < 2,

Absteigend

: 2 < x < \infty

Stecke

x = - 2

\frac{6 x}{x ^{2} + 4} : - \frac{3}{2}

Minimum (- 2, - \frac{3}{2})

Stecke

x = 2

\frac{6 x}{x ^{2} + 4} : \frac{3}{2}

Maximum (2, \frac{3}{2})

Minimum (- 2, - \frac{3}{2}), Maximum (2, \frac{3}{2})

e x t re m e f (x) = 4 - 3 x - x^{2}

e x t re m e f (x) = f (x) = x^{3} - 9 x^{2} + 15 x + 4

e x t re m e f (x) = \frac{x ^{2} + 81}{4 x}, 1 \leq x \leq 12

e x t re m e f (x, y) = 1 + x^{2} - y^{2}

e x t re m e f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 5 x + 2