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

Lösung

extrempunkte

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

Minimum (- \frac{2}{3}, - \frac{16}{3 3} + 5), Maximum (\frac{2}{3}, \frac{16}{3 3} + 5)

Schritte zur Lösung

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

Intervalle finden:

Absteigend

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

Ansteigend

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

Absteigend

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

Stecke

x = - \frac{2}{3}

5 + 4 x - x^{3} : - \frac{16}{3 3} + 5

Minimum (- \frac{2}{3}, - \frac{16}{3 3} + 5)

Stecke

x = \frac{2}{3}

5 + 4 x - x^{3} : \frac{16}{3 3} + 5

Maximum (\frac{2}{3}, \frac{16}{3 3} + 5)

Minimum (- \frac{2}{3}, - \frac{16}{3 3} + 5), Maximum (\frac{2}{3}, \frac{16}{3 3} + 5)

y = (x + 2)^{2}

mi d p o in t (2, - 5), (6, 7)

l in e (2, 4), (3, 7)

s im pl i f y (3.5) (5.3)

a sy m pt o t es f (x) = \frac{x ^{2} - 25}{x + 3}