extreme f(x)=(5x)/(x^2+9)

Lösung

extrempunkte

f (x) = \frac{5 x}{x ^{2} + 9}

Minimum (- 3, - \frac{5}{6}), Maximum (3, \frac{5}{6})

Schritte zur Lösung

f^{'} (x) = \frac{5 ( - x ^{2} + 9 )}{( x ^{2} + 9 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < - 3,

Ansteigend

: - 3 < x < 3,

Absteigend

: 3 < x < \infty

Stecke

x = - 3

\frac{5 x}{x ^{2} + 9} : - \frac{5}{6}

Minimum (- 3, - \frac{5}{6})

Stecke

x = 3

\frac{5 x}{x ^{2} + 9} : \frac{5}{6}

Maximum (3, \frac{5}{6})

Minimum (- 3, - \frac{5}{6}), Maximum (3, \frac{5}{6})

e x t re m e x^{9} - 9 x

e x t re m e x^{2} + 4 x - 5

e x t re m e f (x) = 2 sin (x) - x, 0 \leq x \leq 2 π

e x t re m e f (x) = 8 + 6 x - 7 x^{2} - 4 x^{3}

e x t re m e f (x) = \frac{20}{3 ( 2 + x ) ^{3}}