extreme ln(x^2+x+1)

Lösung

extrempunkte

ln (x^{2} + x + 1)

Minimum (- \frac{1}{2}, ln (\frac{3}{4}))

Schritte zur Lösung

f^{'} (x) = \frac{2 x + 1}{x ^{2} + x + 1}

Intervalle finden:

Absteigend

: - \infty < x < - \frac{1}{2},

Ansteigend

: - \frac{1}{2} < x < \infty

Stecke

x = - \frac{1}{2}

ln (x^{2} + x + 1) : ln (\frac{3}{4})

Minimum (- \frac{1}{2}, ln (\frac{3}{4}))

e x t re m e f (x, y) = 2 x y - x^{2} - 2 y^{2} + 3 x + 4

e x t re m e f (x) = 2 x^{3} - 6 x + 5

e x t re m e (\frac{1}{2} - x^{2} + y^{2}) e^{1 - x^{2} - y^{2}}

e x t re m e f (x) = 2 x ln (9 x)

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