extreme f(x)=x^3-2x^2-15x+1

Lösung

extrempunkte

f (x) = x^{3} - 2 x^{2} - 15 x + 1

Maximum (- \frac{5}{3}, \frac{427}{27}), Minimum (3, - 35)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} - 4 x - 15

Intervalle finden:

Ansteigend

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

Absteigend

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

Ansteigend

: 3 < x < \infty

Stecke

x = - \frac{5}{3}

x^{3} - 2 x^{2} - 15 x + 1 : \frac{427}{27}

Maximum (- \frac{5}{3}, \frac{427}{27})

Stecke

x = 3

x^{3} - 2 x^{2} - 15 x + 1 : - 35

Minimum (3, - 35)

Maximum (- \frac{5}{3}, \frac{427}{27}), Minimum (3, - 35)

e x t re m e f (x) = x^{4} - 242 x^{2} + 7

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

e x t re m e \frac{( x - 2 ) ( 3 - x )}{x ^{2}}

e x t re m e - x^{(2)} - 9

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