inflection-x^4+2x^3+4x^2

Lösung

wendepunkte

- x^{4} + 2 x^{3} + 4 x^{2}

(- \frac{- 3 + 33}{6}, \frac{97 - 15 33}{18}), (\frac{3 + 33}{6}, \frac{97 + 15 33}{18})

Schritte zur Lösung

f^{''} (x) = - 12 x^{2} + 12 x + 8

Intervalle finden:

Konkav fallend

: - \infty < x < - \frac{11}{2 3} + \frac{1}{2},

Konkav steigend

: - \frac{11}{2 3} + \frac{1}{2} < x < \frac{11}{2 3} + \frac{1}{2},

Konkav fallend

: \frac{11}{2 3} + \frac{1}{2} < x < \infty

Stecke

x = - \frac{- 3 + 33}{6}

- x^{4} + 2 x^{3} + 4 x^{2} : \frac{97 - 15 33}{18}

(- \frac{- 3 + 33}{6}, \frac{97 - 15 33}{18})

Stecke

x = \frac{3 + 33}{6}

- x^{4} + 2 x^{3} + 4 x^{2} : \frac{97 + 15 33}{18}

(\frac{3 + 33}{6}, \frac{97 + 15 33}{18})

(- \frac{- 3 + 33}{6}, \frac{97 - 15 33}{18}), (\frac{3 + 33}{6}, \frac{97 + 15 33}{18})

in t erce pt s y = \frac{3}{4} x + 3

d o main f (x) = (8 x - 1) + (- 7 x^{2})

a sy m pt o t es f (x) = \frac{4 x ^{2} - 7 x - 2}{x ^{2} + 1}

mi d p o in t (1, - 1), (5, 3)

l in e m = - 1, (- 5, 1)