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

Lösung

extrempunkte

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

Minimum (- \frac{3}{2}, - \frac{21}{4})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

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

Ansteigend

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

Stecke

x = - \frac{3}{2}

x^{2} + 3 x - 3 : - \frac{21}{4}

Minimum (- \frac{3}{2}, - \frac{21}{4})

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

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

e x t re m e f (x) = \frac{1}{x ^{2} - x - 6}

e x t re m e f (x) = x^{3} - \frac{3}{2} x^{2} - 36 x - 5

e x t re m e f (x, y) = 2 x^{2} - 8 x + y^{2} - 8 y + 7