inflection 3x^4-24x^3+48x^2

Lösung

wendepunkte

3 x^{4} - 24 x^{3} + 48 x^{2}

(\frac{2 ( 3 - 3 )}{3}, 16 (16 - 83) + \frac{64 ( 6 3 - 11 )}{3}), (\frac{2 ( 3 + 3 )}{3}, 16 (16 + 83) - \frac{64 ( 11 + 6 3 )}{3})

Schritte zur Lösung

f^{''} (x) = 36 x^{2} - 144 x + 96

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}

3 x^{4} - 24 x^{3} + 48 x^{2} : 16 (16 - 83) + \frac{64 ( 6 3 - 11 )}{3}

(\frac{2 ( 3 - 3 )}{3}, 16 (16 - 83) + \frac{64 ( 6 3 - 11 )}{3})

Stecke

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

3 x^{4} - 24 x^{3} + 48 x^{2} : 16 (16 + 83) - \frac{64 ( 11 + 6 3 )}{3}

(\frac{2 ( 3 + 3 )}{3}, 16 (16 + 83) - \frac{64 ( 11 + 6 3 )}{3})

(\frac{2 ( 3 - 3 )}{3}, 16 (16 - 83) + \frac{64 ( 6 3 - 11 )}{3}), (\frac{2 ( 3 + 3 )}{3}, 16 (16 + 83) - \frac{64 ( 11 + 6 3 )}{3})

mi d p o in t (6, - 7), (6, 3)

d o main y = \frac{2 x - 3}{12 - ∣ x - 12 ∣}

e x t re m e t^{2} + 2 t - 48

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

d o main f (x) = \frac{4 x - 3}{6 - 3 x}