extreme f(x)=2x^3+x^2-11x

Lösung

extrempunkte

f (x) = 2 x^{3} + x^{2} - 11 x

Maximum (- \frac{1 + 67}{6}, \frac{100 + 67 67}{54}), Minimum (\frac{- 1 + 67}{6}, \frac{100 - 67 67}{54})

Schritte zur Lösung

f^{'} (x) = 6 x^{2} + 2 x - 11

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 67 - 1}{6},

Absteigend

: \frac{- 67 - 1}{6} < x < \frac{67 - 1}{6},

Ansteigend

: \frac{67 - 1}{6} < x < \infty

Stecke

x = - \frac{1 + 67}{6}

2 x^{3} + x^{2} - 11 x : \frac{100 + 67 67}{54}

Maximum (- \frac{1 + 67}{6}, \frac{100 + 67 67}{54})

Stecke

x = \frac{- 1 + 67}{6}

2 x^{3} + x^{2} - 11 x : \frac{100 - 67 67}{54}

Minimum (\frac{- 1 + 67}{6}, \frac{100 - 67 67}{54})

Maximum (- \frac{1 + 67}{6}, \frac{100 + 67 67}{54}), Minimum (\frac{- 1 + 67}{6}, \frac{100 - 67 67}{54})

e x t re m e y = x^{5}

e x t re m e f (x) = 5 y - 5 x + x y - 10

e x t re m e f (x) = ln (x^{2} + 7 x + 15), - 4 \leq x \leq 1

e x t re m e f (x) = - 3 x^{2} + 18 x - 3

e x t re m e f (x) = \frac{( - 2 x + 10 )}{( x - 15 ) ^{11}}