extreme f(x)=100x(2x+3)(x-5)

Lösung

extrempunkte

f (x) = 100 x (2 x + 3) (x - 5)

Maximum (\frac{7 - 139}{6}, \frac{6950 139 - 64400}{27}), Minimum (\frac{7 + 139}{6}, \frac{- 6950 139 - 64400}{27})

Schritte zur Lösung

f^{'} (x) = 100 (6 x^{2} - 14 x - 15)

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 139 + 7}{6},

Absteigend

: \frac{- 139 + 7}{6} < x < \frac{139 + 7}{6},

Ansteigend

: \frac{139 + 7}{6} < x < \infty

Stecke

x = \frac{7 - 139}{6}

100 x (2 x + 3) (x - 5) : \frac{6950 139 - 64400}{27}

Maximum (\frac{7 - 139}{6}, \frac{6950 139 - 64400}{27})

Stecke

x = \frac{7 + 139}{6}

100 x (2 x + 3) (x - 5) : \frac{- 6950 139 - 64400}{27}

Minimum (\frac{7 + 139}{6}, \frac{- 6950 139 - 64400}{27})

Maximum (\frac{7 - 139}{6}, \frac{6950 139 - 64400}{27}), Minimum (\frac{7 + 139}{6}, \frac{- 6950 139 - 64400}{27})

e x t re m e f (x, y) = 9 - x^{2} + 4 - y^{2}

e x t re m e f (x) = x^{3} + 12 x

e x t re m e f (x) = 2 x^{3} - 6 x, 0 \leq x \leq 3

e x t re m e \frac{x ^{4}}{x ^{2} - 1}

e x t re m e f (x, y) = 3 16 - x^{2} - y^{2}