extreme f(x)=sqrt(416-x)x

Lösung

extrempunkte

f (x) = 416 - x x

Maximum (\frac{832}{3}, \frac{3328 26}{3 3}), Minimum (416, 0)

Schritte zur Lösung

f^{'} (x) = \frac{- 3 x + 832}{2 416 - x}

Intervalle finden:

Ansteigend

: - \infty < x < \frac{832}{3},

Absteigend

: \frac{832}{3} < x < 416

Stecke

x = \frac{832}{3}

416 - x x : \frac{3328 26}{3 3}

Maximum (\frac{832}{3}, \frac{3328 26}{3 3})

Stecke

x = 416

416 - x x : 0

Minimum (416, 0)

Maximum (\frac{832}{3}, \frac{3328 26}{3 3}), Minimum (416, 0)

e x t re m e f (x) = x^{\frac{4}{5}} - 4

e x t re m e f (x) = (\frac{5 x ^{4}}{ln ( x )})

e x t re m e f (x) = - (\frac{4}{x ^{2}}), (- 3, - 1)

e x t re m e f (x) = 4 - x^{3} + 5 x

e x t re m e f (x) = 2 x^{3} - 6 x + 11