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

Lösung

extrempunkte

2 x^{3} + 4 x^{2} - 1

Maximum (- \frac{4}{3}, \frac{37}{27}), Minimum (0, - 1)

Schritte zur Lösung

f^{'} (x) = 6 x^{2} + 8 x

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{4}{3},

Absteigend

: - \frac{4}{3} < x < 0,

Ansteigend

: 0 < x < \infty

Stecke

x = - \frac{4}{3}

2 x^{3} + 4 x^{2} - 1 : \frac{37}{27}

Maximum (- \frac{4}{3}, \frac{37}{27})

Stecke

x = 0

2 x^{3} + 4 x^{2} - 1 : - 1

Minimum (0, - 1)

Maximum (- \frac{4}{3}, \frac{37}{27}), Minimum (0, - 1)

e x t re m e f (x) = x \cdot (ln (x))^{5}

e x t re m e f (x, y) = 40000 x + 30000 y - 8 x^{2} - 15 y^{2} - 10 x y

e x t re m e f (x) = e^{x^{3} - 3 x}

e x t re m e f (x) = x + \frac{30}{x}

e x t re m e \frac{5 x + 4}{x + x}