extreme f(x)=20x^3+9x^2-24x+12

Lösung

extrempunkte

f (x) = 20 x^{3} + 9 x^{2} - 24 x + 12

Maximum (- \frac{4}{5}, \frac{668}{25}), Minimum (\frac{1}{2}, \frac{19}{4})

Schritte zur Lösung

f^{'} (x) = 60 x^{2} + 18 x - 24

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{4}{5},

Absteigend

: - \frac{4}{5} < x < \frac{1}{2},

Ansteigend

: \frac{1}{2} < x < \infty

Stecke

x = - \frac{4}{5}

20 x^{3} + 9 x^{2} - 24 x + 12 : \frac{668}{25}

Maximum (- \frac{4}{5}, \frac{668}{25})

Stecke

x = \frac{1}{2}

20 x^{3} + 9 x^{2} - 24 x + 12 : \frac{19}{4}

Minimum (\frac{1}{2}, \frac{19}{4})

Maximum (- \frac{4}{5}, \frac{668}{25}), Minimum (\frac{1}{2}, \frac{19}{4})

e x t re m e f (x) = x^{2} + 9 x + 14

e x t re m e f (x) = - x^{2} - 5 x + 1

e x t re m e f (x) = \frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + 2 x^{2} - 1

e x t re m e - 30 x^{2} - 25 y^{2} + 480 x + 300 y + 170

e x t re m e f (x) = \frac{( x - 2 ) ^{2}}{x - 5}