extreme f(x)=-7x^{5/3}+2\sqrt[3]{x}

Lösung

extrempunkte

f (x) = - 7 x^{\frac{5}{3}} + 2 3 x

Minimum - (\frac{8}{42875})^{\frac{1}{4}}, - 7 (- (\frac{8}{42875})^{\frac{1}{4}})^{\frac{5}{3}} - 2 (\frac{8}{42875})^{\frac{1}{12}}, Maximum (\frac{8}{42875})^{\frac{1}{4}}, - 7 (\frac{8}{42875})^{\frac{5}{12}} + 2 3 (\frac{8}{42875})^{\frac{1}{4}}

Schritte zur Lösung

f^{'} (x) = - \frac{35 x ^{\frac{2}{3}}}{3} + \frac{2}{3 x ^{\frac{2}{3}}}

Intervalle finden:

Absteigend

: - \infty < x < - (\frac{2}{35})^{\frac{3}{4}},

Ansteigend

: - (\frac{2}{35})^{\frac{3}{4}} < x < 0,

Ansteigend

: 0 < x < (\frac{2}{35})^{\frac{3}{4}},

Absteigend

: (\frac{2}{35})^{\frac{3}{4}} < x < \infty

Stecke

x = - (\frac{8}{42875})^{\frac{1}{4}}

- 7 x^{\frac{5}{3}} + 2 3 x : - 7 (- (\frac{8}{42875})^{\frac{1}{4}})^{\frac{5}{3}} - 2 (\frac{8}{42875})^{\frac{1}{12}}

Minimum - (\frac{8}{42875})^{\frac{1}{4}}, - 7 (- (\frac{8}{42875})^{\frac{1}{4}})^{\frac{5}{3}} - 2 (\frac{8}{42875})^{\frac{1}{12}}

Stecke

x = (\frac{8}{42875})^{\frac{1}{4}}

- 7 x^{\frac{5}{3}} + 2 3 x : - 7 (\frac{8}{42875})^{\frac{5}{12}} + 2 3 (\frac{8}{42875})^{\frac{1}{4}}

Maximum (\frac{8}{42875})^{\frac{1}{4}}, - 7 (\frac{8}{42875})^{\frac{5}{12}} + 2 3 (\frac{8}{42875})^{\frac{1}{4}}

Minimum - (\frac{8}{42875})^{\frac{1}{4}}, - 7 (- (\frac{8}{42875})^{\frac{1}{4}})^{\frac{5}{3}} - 2 (\frac{8}{42875})^{\frac{1}{12}}, Maximum (\frac{8}{42875})^{\frac{1}{4}}, - 7 (\frac{8}{42875})^{\frac{5}{12}} + 2 3 (\frac{8}{42875})^{\frac{1}{4}}

e x t re m e f (x, y) = 2 x^{2} - 4 y + y^{2}

e x t re m e f (x) = 3 x + 4 32.5 - x^{2}

e x t re m e f (x) = \frac{x + 6}{x - 3}

e x t re m e f (x) = 2 x^{3} + 3 x^{2} - 72 x + 1, - 4 \leq x \leq 7

e x t re m e f (x) = \frac{( x ^{2} )}{x ^{2} + 243}