extreme f(x)=2x^3+3x^2-2x-3

Lösung

extrempunkte

f (x) = 2 x^{3} + 3 x^{2} - 2 x - 3

Maximum (- \frac{3 + 21}{6}, \frac{27 + 7 21}{18} - 3), Minimum (\frac{- 3 + 21}{6}, \frac{27 - 7 21}{18} - 3)

Schritte zur Lösung

f^{'} (x) = 6 x^{2} + 6 x - 2

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{7}{2 3} - \frac{1}{2},

Absteigend

: - \frac{7}{2 3} - \frac{1}{2} < x < \frac{7}{2 3} - \frac{1}{2},

Ansteigend

: \frac{7}{2 3} - \frac{1}{2} < x < \infty

Stecke

x = - \frac{3 + 21}{6}

2 x^{3} + 3 x^{2} - 2 x - 3 : \frac{27 + 7 21}{18} - 3

Maximum (- \frac{3 + 21}{6}, \frac{27 + 7 21}{18} - 3)

Stecke

x = \frac{- 3 + 21}{6}

2 x^{3} + 3 x^{2} - 2 x - 3 : \frac{27 - 7 21}{18} - 3

Minimum (\frac{- 3 + 21}{6}, \frac{27 - 7 21}{18} - 3)

Maximum (- \frac{3 + 21}{6}, \frac{27 + 7 21}{18} - 3), Minimum (\frac{- 3 + 21}{6}, \frac{27 - 7 21}{18} - 3)

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

e x t re m e f (x) = \frac{x ^{2} + 2}{x ^{2} - 9}

e x t re m e f (x) = \frac{x ^{3}}{3} - 5 x^{2} - 50 x + 1

e x t re m e \frac{4}{x} + x^{4}

e x t re m e f (x, y) = 9 x^{2} - 4 y^{2}