extreme f(x)= 1/5 x^3-sqrt(x^2+1)

Lösung

extrempunkte

f (x) = \frac{1}{5} x^{3} - x^{2} + 1

Maximum (0, - 1), Minimum \frac{- 3 + 109}{6}, \frac{( - 3 + 109 ) ^{\frac{3}{2}}}{30 6} - \frac{- 3 + 109}{6} + 1

Schritte zur Lösung

f^{'} (x) = \frac{3 x ^{2}}{5} - \frac{x}{x ^{2} + 1}

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Absteigend

: 0 < x < \frac{109 - 3}{6},

Ansteigend

: \frac{109 - 3}{6} < x < \infty

Stecke

x = 0

\frac{1}{5} x^{3} - x^{2} + 1 : - 1

Maximum (0, - 1)

Stecke

x = \frac{- 3 + 109}{6}

\frac{1}{5} x^{3} - x^{2} + 1 : \frac{( - 3 + 109 ) ^{\frac{3}{2}}}{30 6} - \frac{- 3 + 109}{6} + 1

Minimum \frac{- 3 + 109}{6}, \frac{( - 3 + 109 ) ^{\frac{3}{2}}}{30 6} - \frac{- 3 + 109}{6} + 1

Maximum (0, - 1), Minimum \frac{- 3 + 109}{6}, \frac{( - 3 + 109 ) ^{\frac{3}{2}}}{30 6} - \frac{- 3 + 109}{6} + 1

e x t re m e \frac{4 x}{x ^{2} - 1}

e x t re m e 9 - x^{2}

e x t re m e f (x) = 3 x^{3} + 3 x^{2}

e x t re m e x^{2} (x - 5) (x^{2} + 4 x + 8)

e x t re m e f (x) = - 2 x^{2} + 3 x + 5