extreme f(x)=x^3+x^2-46x+80

Lösung

extrempunkte

f (x) = x^{3} + x^{2} - 46 x + 80

Maximum (- \frac{1 + 139}{3}, \frac{416 + 278 139}{27} + 80), Minimum (\frac{- 1 + 139}{3}, \frac{416 - 278 139}{27} + 80)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} + 2 x - 46

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 139 - 1}{3},

Absteigend

: \frac{- 139 - 1}{3} < x < \frac{139 - 1}{3},

Ansteigend

: \frac{139 - 1}{3} < x < \infty

Stecke

x = - \frac{1 + 139}{3}

x^{3} + x^{2} - 46 x + 80 : \frac{416 + 278 139}{27} + 80

Maximum (- \frac{1 + 139}{3}, \frac{416 + 278 139}{27} + 80)

Stecke

x = \frac{- 1 + 139}{3}

x^{3} + x^{2} - 46 x + 80 : \frac{416 - 278 139}{27} + 80

Minimum (\frac{- 1 + 139}{3}, \frac{416 - 278 139}{27} + 80)

Maximum (- \frac{1 + 139}{3}, \frac{416 + 278 139}{27} + 80), Minimum (\frac{- 1 + 139}{3}, \frac{416 - 278 139}{27} + 80)

e x t re m e f (x) = 3 x^{4} + 8 x^{3} - 6 x^{2} + 24 x

e x t re m e 3 x (x - 3)^{3}

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

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

e x t re m e - \frac{1}{80} t^{2} + \frac{9}{40} t - \frac{17}{80}