inflection f(x)=(x^2-4x)^2

Lösung

wendepunkte

f (x) = (x^{2} - 4 x)^{2}

(\frac{2 ( 3 - 3 )}{3}, \frac{64}{9}), (\frac{2 ( 3 + 3 )}{3}, \frac{64}{9})

Schritte zur Lösung

f^{''} (x) = 2 (6 x^{2} - 24 x + 16)

Intervalle finden:

Konkav steigend

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

Konkav fallend

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

Konkav steigend

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

Stecke

x = \frac{2 ( 3 - 3 )}{3}

(x^{2} - 4 x)^{2} : \frac{64}{9}

(\frac{2 ( 3 - 3 )}{3}, \frac{64}{9})

Stecke

x = \frac{2 ( 3 + 3 )}{3}

(x^{2} - 4 x)^{2} : \frac{64}{9}

(\frac{2 ( 3 + 3 )}{3}, \frac{64}{9})

(\frac{2 ( 3 - 3 )}{3}, \frac{64}{9}), (\frac{2 ( 3 + 3 )}{3}, \frac{64}{9})

s l o p e 3 x + 7 y = 4 y - 2

r an g e \frac{x}{2 x ^{2} + 9 x + 4}

s l o p e in t erce pt - 2 x + y = 4

in f l ec t i o n f (x) = x^{7} + 7 x^{3} + 1

p er p e n d i c u l a r y = - 4 x - 6, (9, 4)