extreme f(x)=x^4-7x^2+8

Lösung

extrempunkte

f (x) = x^{4} - 7 x^{2} + 8

Minimum (- \frac{7}{2}, - \frac{17}{4}), Maximum (0, 8), Minimum (\frac{7}{2}, - \frac{17}{4})

Schritte zur Lösung

f^{'} (x) = 4 x^{3} - 14 x

Intervalle finden:

Absteigend

: - \infty < x < - \frac{7}{2},

Ansteigend

: - \frac{7}{2} < x < 0,

Absteigend

: 0 < x < \frac{7}{2},

Ansteigend

: \frac{7}{2} < x < \infty

Stecke

x = - \frac{7}{2}

x^{4} - 7 x^{2} + 8 : - \frac{17}{4}

Minimum (- \frac{7}{2}, - \frac{17}{4})

Stecke

x = 0

x^{4} - 7 x^{2} + 8 : 8

Maximum (0, 8)

Stecke

x = \frac{7}{2}

x^{4} - 7 x^{2} + 8 : - \frac{17}{4}

Minimum (\frac{7}{2}, - \frac{17}{4})

Minimum (- \frac{7}{2}, - \frac{17}{4}), Maximum (0, 8), Minimum (\frac{7}{2}, - \frac{17}{4})

s l o p e in t erce pt y - 3 = \frac{5}{3} (x - 6)

d o main f (x) = \frac{6}{16 - x ^{2}}

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

d o main f (x) = - (x + 1) (x - 2) (x - 3)

in v erse f (x) = 7 x^{3} - 3