extreme f(x)=2x^3+4x^2-1

Lösung

extrempunkte

f (x) = 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)

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e x t re m e Z (x, y) = x + y - (\frac{x \cdot y}{100})

e x t re m e f (x, y) = x^{2} + y^{2} + 12 x - 16 y

e x t re m e f (x, y) = 49 - (x - 9)^{2} - y^{2}

e x t re m e f (x) = 4 x^{2} - 5 x - 2