extreme y=x^2(x^2-5)

Lösung

extrempunkte

y = x^{2} (x^{2} - 5)

Minimum (- \frac{5}{2}, - \frac{25}{4}), Maximum (0, 0), Minimum (\frac{5}{2}, - \frac{25}{4})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < - \frac{5}{2},

Ansteigend

: - \frac{5}{2} < x < 0,

Absteigend

: 0 < x < \frac{5}{2},

Ansteigend

: \frac{5}{2} < x < \infty

Stecke

x = - \frac{5}{2}

x^{2} (x^{2} - 5) : - \frac{25}{4}

Minimum (- \frac{5}{2}, - \frac{25}{4})

Stecke

x = 0

x^{2} (x^{2} - 5) : 0

Maximum (0, 0)

Stecke

x = \frac{5}{2}

x^{2} (x^{2} - 5) : - \frac{25}{4}

Minimum (\frac{5}{2}, - \frac{25}{4})

Minimum (- \frac{5}{2}, - \frac{25}{4}), Maximum (0, 0), Minimum (\frac{5}{2}, - \frac{25}{4})

e x t re m e f (x) = 6 x + 18

e x t re m e f (x) = \frac{4 y + 7}{3 - x}

e x t re m e 3 cos^{2} (x), 0 \leq x \leq π

e x t re m e f (x) = x + \frac{9}{x} + 2, 1 \leq x \leq 18

e x t re m e f (x) = 7 x^{4} - 6 x^{2} - 1