extreme f(x)=sqrt(x+12)-8

Lösung

extrempunkte

f (x) = x + 12 - 8

Minimum (- 12, - 8)

Schritte zur Lösung

f^{'} (x) = \frac{1}{2 x + 12}

Intervalle finden:

Ansteigend

: - 12 < x < \infty

Stecke

x = - 12

x + 12 - 8 : - 8

Minimum (- 12, - 8)

e x t re m e f (x, y) = 4 x y + 120 y - 20 y^{2} - \frac{1}{10} x^{2} y - 90

e x t re m e f (x, y) = 2 x^{2} + 4 y^{2} - 2 x y - 10 x - 2 y + 2

e x t re m e f (x) = \frac{7 x}{x ^{2} + 9}, 0 \leq x \leq 9

e x t re m e 2 x^{3} + 4 x^{2} - 1

e x t re m e f (x) = x \cdot (ln (x))^{5}