extreme f(x)= 3/5 x^3-7x+4

Lösung

extrempunkte

f (x) = \frac{3}{5} x^{3} - 7 x + 4

Maximum (- \frac{35}{3}, \frac{14 35}{9} + 4), Minimum (\frac{35}{3}, - \frac{14 35}{9} + 4)

Schritte zur Lösung

f^{'} (x) = \frac{9 x ^{2}}{5} - 7

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{35}{3},

Absteigend

: - \frac{35}{3} < x < \frac{35}{3},

Ansteigend

: \frac{35}{3} < x < \infty

Stecke

x = - \frac{35}{3}

\frac{3}{5} x^{3} - 7 x + 4 : \frac{14 35}{9} + 4

Maximum (- \frac{35}{3}, \frac{14 35}{9} + 4)

Stecke

x = \frac{35}{3}

\frac{3}{5} x^{3} - 7 x + 4 : - \frac{14 35}{9} + 4

Minimum (\frac{35}{3}, - \frac{14 35}{9} + 4)

Maximum (- \frac{35}{3}, \frac{14 35}{9} + 4), Minimum (\frac{35}{3}, - \frac{14 35}{9} + 4)

in v erse f (x) = \frac{x + 6}{x - 5}

a sy m pt o t es f (x) = tan (\frac{π}{3} x)

in v erse \frac{x + 2}{x - 2}

r an g e f (x) = x - 5

in v erse ln (7 x)