extreme f(x)=x^2+x+1

Lösung

extrempunkte

f (x) = x^{2} + x + 1

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

Schritte zur Lösung

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

Intervalle finden:

Absteigend

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

Ansteigend

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

Stecke

x = - \frac{1}{2}

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

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

e x t re m e f (x) = 7 x^{\frac{2}{5}} + 2 x^{\frac{7}{5}}

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

e x t re m e f (x) = (\frac{4860}{x}) + 12 x + 709407

e x t re m e \frac{10000}{2.5 cos ( x ) + sin ( x )}

e x t re m e f (x) = (x - 9) e^{x}