extreme f(x)=6x^3-5x+12

Lösung

extrempunkte

f (x) = 6 x^{3} - 5 x + 12

Maximum (- \frac{5}{3 2}, \frac{5 10}{9} + 12), Minimum (\frac{5}{3 2}, - \frac{5 10}{9} + 12)

Schritte zur Lösung

f^{'} (x) = 18 x^{2} - 5

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{5}{3 2},

Absteigend

: - \frac{5}{3 2} < x < \frac{5}{3 2},

Ansteigend

: \frac{5}{3 2} < x < \infty

Stecke

x = - \frac{5}{3 2}

6 x^{3} - 5 x + 12 : \frac{5 10}{9} + 12

Maximum (- \frac{5}{3 2}, \frac{5 10}{9} + 12)

Stecke

x = \frac{5}{3 2}

6 x^{3} - 5 x + 12 : - \frac{5 10}{9} + 12

Minimum (\frac{5}{3 2}, - \frac{5 10}{9} + 12)

Maximum (- \frac{5}{3 2}, \frac{5 10}{9} + 12), Minimum (\frac{5}{3 2}, - \frac{5 10}{9} + 12)

in t erce pt s 4 x^{2} + 8 x - 3

a sy m pt o t es y = \frac{x ^{2} - x}{x ^{2} - 8 x + 7}

s l o p e 3 y + x = 12

d o main f (x) = - x + 10

d o main f (x) = 3 x - \frac{2}{x + 1}