extreme 1/3 x^3+2x^2+4x+1

Lösung

extrempunkte

\frac{1}{3} x^{3} + 2 x^{2} + 4 x + 1

Sattel (- 2, - \frac{5}{3})

Schritte zur Lösung

f^{'} (x) = x^{2} + 4 x + 4

Intervalle finden:

Ansteigend

: - \infty < x < - 2,

Ansteigend

: - 2 < x < \infty

Stecke

x = - 2

\frac{1}{3} x^{3} + 2 x^{2} + 4 x + 1 : - \frac{5}{3}

Sattel (- 2, - \frac{5}{3})

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

e x t re m e f (x, y) = 3 x^{(4)} y^{(5)} + 6 x - 7 y

e x t re m e \frac{4 x}{x ^{2} + 4}

e x t re m e f (x) = - 0.018 x^{3} + 0.215 x^{2} + 0.057 x + 1

e x t re m e f (x) = 2 x^{3} + 9 x^{2} - 60 x + 7, - 5 \leq x \leq 5