extreme x^3-4x^2+3x

Lösung

extrempunkte

x^{3} - 4 x^{2} + 3 x

Maximum (\frac{4 - 7}{3}, - 7 + \frac{41 7 - 128}{27} + 4), Minimum (\frac{4 + 7}{3}, 7 + \frac{- 128 - 41 7}{27} + 4)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 7 + 4}{3},

Absteigend

: \frac{- 7 + 4}{3} < x < \frac{7 + 4}{3},

Ansteigend

: \frac{7 + 4}{3} < x < \infty

Stecke

x = \frac{4 - 7}{3}

x^{3} - 4 x^{2} + 3 x : - 7 + \frac{41 7 - 128}{27} + 4

Maximum (\frac{4 - 7}{3}, - 7 + \frac{41 7 - 128}{27} + 4)

Stecke

x = \frac{4 + 7}{3}

x^{3} - 4 x^{2} + 3 x : 7 + \frac{- 128 - 41 7}{27} + 4

Minimum (\frac{4 + 7}{3}, 7 + \frac{- 128 - 41 7}{27} + 4)

Maximum (\frac{4 - 7}{3}, - 7 + \frac{41 7 - 128}{27} + 4), Minimum (\frac{4 + 7}{3}, 7 + \frac{- 128 - 41 7}{27} + 4)

e x t re m e f (x, y) = 3 x^{2} + 4 x y - 4 y^{2} - x + 3 y

e x t re m e \frac{( x + 1 ) ^{2}}{1 + x ^{2}}

e x t re m e f (x) = x (17 - 43 + 2 x) (\frac{43}{2} - x)

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

e x t re m e ln^{2} (x)