extreme f(x)=(x^2)/(x^4+16)

Lösung

extrempunkte

f (x) = \frac{x ^{2}}{x ^{4} + 16}

Maximum (- 2, \frac{1}{8}), Minimum (0, 0), Maximum (2, \frac{1}{8})

Schritte zur Lösung

f^{'} (x) = \frac{- 2 x ^{5} + 32 x}{( x ^{4} + 16 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < - 2,

Absteigend

: - 2 < x < 0,

Ansteigend

: 0 < x < 2,

Absteigend

: 2 < x < \infty

Stecke

x = - 2

\frac{x ^{2}}{x ^{4} + 16} : \frac{1}{8}

Maximum (- 2, \frac{1}{8})

Stecke

x = 0

\frac{x ^{2}}{x ^{4} + 16} : 0

Minimum (0, 0)

Stecke

x = 2

\frac{x ^{2}}{x ^{4} + 16} : \frac{1}{8}

Maximum (2, \frac{1}{8})

Maximum (- 2, \frac{1}{8}), Minimum (0, 0), Maximum (2, \frac{1}{8})

e x t re m e (x - 2)^{3} \cdot (x + 2)

e x t re m e f (x) = \frac{x + 2}{x ^{2} - 7 x - 18}

e x t re m e 5 x y - 7 x^{2} + 3 x - 6 y + 2

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

e x t re m e f (x) = 4 x^{3} - 48 x - 1