extreme f(x)=0.0001x^2+3.2x-90

Lösung

extrempunkte

f (x) = 0.0001 x^{2} + 3.2 x - 90

Minimum (- \frac{3.2}{0.0002}, - 25690)

Schritte zur Lösung

f^{'} (x) = 0.0002 x + 3.2

Intervalle finden:

Absteigend

: - \infty < x < - \frac{3.2}{0.0002},

Ansteigend

: - \frac{3.2}{0.0002} < x < \infty

Stecke

x = - \frac{3.2}{0.0002}

0.0001 x^{2} + 3.2 x - 90 : - 25690

Minimum (- \frac{3.2}{0.0002}, - 25690)

e x t re m e f (x, y) = θ

e x t re m e \frac{x ^{3}}{3} - x^{2} - 35 x + 4

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e x t re m e f (x, y) = x y - 8 x - y^{2} + 12 x + 160

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