extreme (x^3)/3-x^2-35x+4

Lösung

extrempunkte

\frac{x ^{3}}{3} - x^{2} - 35 x + 4

Maximum (- 5, \frac{337}{3}), Minimum (7, - \frac{527}{3})

Schritte zur Lösung

f^{'} (x) = x^{2} - 2 x - 35

Intervalle finden:

Ansteigend

: - \infty < x < - 5,

Absteigend

: - 5 < x < 7,

Ansteigend

: 7 < x < \infty

Stecke

x = - 5

\frac{x ^{3}}{3} - x^{2} - 35 x + 4 : \frac{337}{3}

Maximum (- 5, \frac{337}{3})

Stecke

x = 7

\frac{x ^{3}}{3} - x^{2} - 35 x + 4 : - \frac{527}{3}

Minimum (7, - \frac{527}{3})

Maximum (- 5, \frac{337}{3}), Minimum (7, - \frac{527}{3})

e x t re m e f (x, y) = e^{6 x^{2} + 4 y^{2} + 10}

e x t re m e f (x, y) = x y - 8 x - y^{2} + 12 x + 160

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

e x t re m e f (x) = 0.4 x^{2} + 110.2 x, 0 \leq x

e x t re m e f (x) = \frac{( 3 x )}{( 9 - x ^{2} )}, - 3 \leq x \leq 2