extreme f(x)=sqrt(6x^3+8x^2)

Lösung

extrempunkte

f (x) = 6 x^{3} + 8 x^{2}

Minimum (- \frac{4}{3}, 0), Maximum (- \frac{8}{9}, \frac{16 2}{9 3}), Minimum (0, 0)

Schritte zur Lösung

f^{'} (x) = \frac{x ( 9 x + 8 )}{2 x ^{2} 3 x + 4}

Intervalle finden:

Ansteigend

: - \frac{4}{3} < x < - \frac{8}{9},

Absteigend

: - \frac{8}{9} < x < 0,

Ansteigend

: 0 < x < \infty

Stecke

x = - \frac{4}{3}

6 x^{3} + 8 x^{2} : 0

Minimum (- \frac{4}{3}, 0)

Stecke

x = - \frac{8}{9}

6 x^{3} + 8 x^{2} : \frac{16 2}{9 3}

Maximum (- \frac{8}{9}, \frac{16 2}{9 3})

Stecke

x = 0

6 x^{3} + 8 x^{2} : 0

Minimum (0, 0)

Minimum (- \frac{4}{3}, 0), Maximum (- \frac{8}{9}, \frac{16 2}{9 3}), Minimum (0, 0)

in t erce pt s f (x) = - 5 x^{2} - 30 x - 42

l in e (- 1, 4), (1, - 6)

a sy m pt o t es f (x) = \frac{x ^{3} + 5}{x ^{5} + 2}

r an g e f (x) = - \frac{1}{3} x

in t erce pt s f (x) = 4 x^{4} + 12 x^{3} - 40 x^{2}