extreme f(x)=x^3+4x^2-3x+6

Lösung

extrempunkte

f (x) = x^{3} + 4 x^{2} - 3 x + 6

Maximum (- 3, 24), Minimum (\frac{1}{3}, \frac{148}{27})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 3,

Absteigend

: - 3 < x < \frac{1}{3},

Ansteigend

: \frac{1}{3} < x < \infty

Stecke

x = - 3

x^{3} + 4 x^{2} - 3 x + 6 : 24

Maximum (- 3, 24)

Stecke

x = \frac{1}{3}

x^{3} + 4 x^{2} - 3 x + 6 : \frac{148}{27}

Minimum (\frac{1}{3}, \frac{148}{27})

Maximum (- 3, 24), Minimum (\frac{1}{3}, \frac{148}{27})

e x t re m e f (x, y) = x^{3} + y^{3} + 2 x^{2} + 4 y^{2} + 6

e x t re m e f (x, y) = 5 + 3 x^{2} + 3 y^{2} + 2 y^{3} + x^{3}

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

e x t re m e x^{\frac{2}{3}} - 6 x^{- \frac{1}{3}}

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