extreme f(x)=sqrt(x^2+9)

Lösung

extrempunkte

f (x) = x^{2} + 9

Minimum (0, 3)

Schritte zur Lösung

f^{'} (x) = \frac{x}{x ^{2} + 9}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Ansteigend

: 0 < x < \infty

Stecke

x = 0

x^{2} + 9 : 3

Minimum (0, 3)

e x t re m e \frac{4}{x + 6}

e x t re m e f (x, y) = 113 x + 15 y - x^{2} - 2 y^{2} + 3 x y

e x t re m e \frac{csc ( x )}{1 + csc ( x )}

e x t re m e f (x) = x^{\frac{8}{3}}, - 8 \leq x \leq 1

e x t re m e f (x) = cos^{2} (x), 0 \leq x \leq π