extreme f(x)=x^3+2x^2-x+8

Lösung

extrempunkte

f (x) = x^{3} + 2 x^{2} - x + 8

Maximum (- \frac{2 + 7}{3}, \frac{34 + 14 7}{27} + 8), Minimum (\frac{- 2 + 7}{3}, \frac{34 - 14 7}{27} + 8)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} + 4 x - 1

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 7 - 2}{3},

Absteigend

: \frac{- 7 - 2}{3} < x < \frac{7 - 2}{3},

Ansteigend

: \frac{7 - 2}{3} < x < \infty

Stecke

x = - \frac{2 + 7}{3}

x^{3} + 2 x^{2} - x + 8 : \frac{34 + 14 7}{27} + 8

Maximum (- \frac{2 + 7}{3}, \frac{34 + 14 7}{27} + 8)

Stecke

x = \frac{- 2 + 7}{3}

x^{3} + 2 x^{2} - x + 8 : \frac{34 - 14 7}{27} + 8

Minimum (\frac{- 2 + 7}{3}, \frac{34 - 14 7}{27} + 8)

Maximum (- \frac{2 + 7}{3}, \frac{34 + 14 7}{27} + 8), Minimum (\frac{- 2 + 7}{3}, \frac{34 - 14 7}{27} + 8)

e x t re m e f (x) = \frac{1}{3} x^{3} - 4 x^{2} + 12 x - 5

e x t re m e f (x) = x 2 - x

e x t re m e f (x, y, z) = x + y 2 + x 3

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

e x t re m e f (x) = ln (5 - 2 x^{2})