extreme f(x)=-2x^{5/3}+4\sqrt[3]{x}

Lösung

extrempunkte

f (x) = - 2 x^{\frac{5}{3}} + 4 3 x

Minimum - (\frac{8}{125})^{\frac{1}{4}}, - 2 (- (\frac{8}{125})^{\frac{1}{4}})^{\frac{5}{3}} - 4 (\frac{8}{125})^{\frac{1}{12}}, Maximum (\frac{8}{125})^{\frac{1}{4}}, - 2 (\frac{8}{125})^{\frac{5}{12}} + 4 3 (\frac{8}{125})^{\frac{1}{4}}

Schritte zur Lösung

f^{'} (x) = - \frac{10 x ^{\frac{2}{3}}}{3} + \frac{4}{3 x ^{\frac{2}{3}}}

Intervalle finden:

Absteigend

: - \infty < x < - (\frac{2}{5})^{\frac{3}{4}},

Ansteigend

: - (\frac{2}{5})^{\frac{3}{4}} < x < 0,

Ansteigend

: 0 < x < (\frac{2}{5})^{\frac{3}{4}},

Absteigend

: (\frac{2}{5})^{\frac{3}{4}} < x < \infty

Stecke

x = - (\frac{8}{125})^{\frac{1}{4}}

- 2 x^{\frac{5}{3}} + 4 3 x : - 2 (- (\frac{8}{125})^{\frac{1}{4}})^{\frac{5}{3}} - 4 (\frac{8}{125})^{\frac{1}{12}}

Minimum - (\frac{8}{125})^{\frac{1}{4}}, - 2 (- (\frac{8}{125})^{\frac{1}{4}})^{\frac{5}{3}} - 4 (\frac{8}{125})^{\frac{1}{12}}

Stecke

x = (\frac{8}{125})^{\frac{1}{4}}

- 2 x^{\frac{5}{3}} + 4 3 x : - 2 (\frac{8}{125})^{\frac{5}{12}} + 4 3 (\frac{8}{125})^{\frac{1}{4}}

Maximum (\frac{8}{125})^{\frac{1}{4}}, - 2 (\frac{8}{125})^{\frac{5}{12}} + 4 3 (\frac{8}{125})^{\frac{1}{4}}

Minimum - (\frac{8}{125})^{\frac{1}{4}}, - 2 (- (\frac{8}{125})^{\frac{1}{4}})^{\frac{5}{3}} - 4 (\frac{8}{125})^{\frac{1}{12}}, Maximum (\frac{8}{125})^{\frac{1}{4}}, - 2 (\frac{8}{125})^{\frac{5}{12}} + 4 3 (\frac{8}{125})^{\frac{1}{4}}

e x t re m e f (x) = \frac{( x ^{2} - 5 x + 4 )}{x - 5}

e x t re m e - x^{3} - 9 x^{2}

e x t re m e y = \frac{x ^{2} + 2}{x ^{2} - 25}

e x t re m e f (x) = - sin (x) - 4, (0, 2 π)

e x t re m e f (x) = 6 + 9 x^{2} - 6 x^{3}