inflection f(x)=x^2e^{4x}

Lösung

wendepunkte

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

(- \frac{2 + 1}{2 2}, \frac{e ^{- 2 - 2} ( 3 + 2 2 )}{8}), (\frac{- 2 + 2}{4}, \frac{e ^{- 2 + 2} ( 3 - 2 2 )}{8})

Schritte zur Lösung

f^{''} (x) = 2 e^{4 x} (1 + 8 x + 8 x^{2})

Intervalle finden:

Konkav steigend

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

Konkav fallend

: - \frac{1}{2 2} - \frac{1}{2} < x < \frac{1}{2 2} - \frac{1}{2},

Konkav steigend

: \frac{1}{2 2} - \frac{1}{2} < x < \infty

Stecke

x = - \frac{2 + 1}{2 2}

x^{2} e^{4 x} : \frac{e ^{- 2 - 2} ( 3 + 2 2 )}{8}

(- \frac{2 + 1}{2 2}, \frac{e ^{- 2 - 2} ( 3 + 2 2 )}{8})

Stecke

x = \frac{- 2 + 2}{4}

x^{2} e^{4 x} : \frac{e ^{- 2 + 2} ( 3 - 2 2 )}{8}

(\frac{- 2 + 2}{4}, \frac{e ^{- 2 + 2} ( 3 - 2 2 )}{8})

(- \frac{2 + 1}{2 2}, \frac{e ^{- 2 - 2} ( 3 + 2 2 )}{8}), (\frac{- 2 + 2}{4}, \frac{e ^{- 2 + 2} ( 3 - 2 2 )}{8})

d o main \frac{2 x ^{3} - 5}{x ^{2} + x - 6}

in v erse f (x) = - x^{2} - 4 x + 1

in v erse f (x) = \frac{( 2 x - 3 )}{4 x + 5}

cr i t i c a l x^{2} + 2 x - 15

e x t re m e y = \frac{1}{x} + x