extreme f(x)=4+5x+x^2

Lösung

extrempunkte

f (x) = 4 + 5 x + x^{2}

Minimum (- \frac{5}{2}, - \frac{9}{4})

Schritte zur Lösung

f^{'} (x) = 2 x + 5

Intervalle finden:

Absteigend

: - \infty < x < - \frac{5}{2},

Ansteigend

: - \frac{5}{2} < x < \infty

Stecke

x = - \frac{5}{2}

4 + 5 x + x^{2} : - \frac{9}{4}

Minimum (- \frac{5}{2}, - \frac{9}{4})

e x t re m e - 2 x + 3 ln (4 x), 1 \leq x \leq 5

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

e x t re m e f (x, y) = x^{2} + x y + y^{2} - 9 x + 1

e x t re m e \frac{4 s}{s ^{2} - 16}

e x t re m e f (x, y) = 107 - 4 x^{2} - 3 y^{2}