extreme y=x^3+4x^2+5x-2

Lösung

extrempunkte

y = x^{3} + 4 x^{2} + 5 x - 2

Maximum (- \frac{5}{3}, - \frac{104}{27}), Minimum (- 1, - 4)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} + 8 x + 5

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{5}{3},

Absteigend

: - \frac{5}{3} < x < - 1,

Ansteigend

: - 1 < x < \infty

Stecke

x = - \frac{5}{3}

x^{3} + 4 x^{2} + 5 x - 2 : - \frac{104}{27}

Maximum (- \frac{5}{3}, - \frac{104}{27})

Stecke

x = - 1

x^{3} + 4 x^{2} + 5 x - 2 : - 4

Minimum (- 1, - 4)

Maximum (- \frac{5}{3}, - \frac{104}{27}), Minimum (- 1, - 4)

e x t re m e \frac{x - 4}{x ^{2}}

e x t re m e f (x) = x^{4} + 6 x^{5} - x

e x t re m e f (x) = - 2 x^{2} + 8 x - 1

e x t re m e f (x) = x^{3} - 12 x - 8

e x t re m e f (x) = 7 x - 2 x ln (x)