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

Lösung

extrempunkte

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

Maximum (- \frac{2 + 13}{3}, 13 + \frac{16 - 13}{27} + 2), Minimum (\frac{- 2 + 13}{3}, 2 - 13 + \frac{16 + 13}{27})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 13 - 2}{3},

Absteigend

: \frac{- 13 - 2}{3} < x < \frac{13 - 2}{3},

Ansteigend

: \frac{13 - 2}{3} < x < \infty

Stecke

x = - \frac{2 + 13}{3}

x^{3} + 2 x^{2} - 3 x : 13 + \frac{16 - 13}{27} + 2

Maximum (- \frac{2 + 13}{3}, 13 + \frac{16 - 13}{27} + 2)

Stecke

x = \frac{- 2 + 13}{3}

x^{3} + 2 x^{2} - 3 x : 2 - 13 + \frac{16 + 13}{27}

Minimum (\frac{- 2 + 13}{3}, 2 - 13 + \frac{16 + 13}{27})

Maximum (- \frac{2 + 13}{3}, 13 + \frac{16 - 13}{27} + 2), Minimum (\frac{- 2 + 13}{3}, 2 - 13 + \frac{16 + 13}{27})

e x t re m e - 5 x^{3} + 2 x^{2} - x + 20

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

e x t re m e P (a, b) = a^{2} - 6 a + 9 - b^{2}

e x t re m e f (x) = x ln (\frac{x}{3}), 0.1 \leq x \leq 3

e x t re m e f (x) = 2 x^{3} - 6 x^{2} + 6 x - 8