extreme f(x)=2x^{5/3}-3x^{2/3}

Lösung

extrempunkte

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

Maximum (0, 0), Minimum (\frac{3}{5}, 2 (\frac{3}{5})^{\frac{5}{3}} - 3 (\frac{3}{5})^{\frac{2}{3}})

Schritte zur Lösung

f^{'} (x) = \frac{10 x - 6}{3 x ^{\frac{1}{3}}}

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Absteigend

: 0 < x < \frac{3}{5},

Ansteigend

: \frac{3}{5} < x < \infty

Stecke

x = 0

2 x^{\frac{5}{3}} - 3 x^{\frac{2}{3}} : 0

Maximum (0, 0)

Stecke

x = \frac{3}{5}

2 x^{\frac{5}{3}} - 3 x^{\frac{2}{3}} : 2 (\frac{3}{5})^{\frac{5}{3}} - 3 (\frac{3}{5})^{\frac{2}{3}}

Minimum (\frac{3}{5}, 2 (\frac{3}{5})^{\frac{5}{3}} - 3 (\frac{3}{5})^{\frac{2}{3}})

Maximum (0, 0), Minimum (\frac{3}{5}, 2 (\frac{3}{5})^{\frac{5}{3}} - 3 (\frac{3}{5})^{\frac{2}{3}})

e x t re m e y = sin (x) cos (x), 0 \leq x \leq π

e x t re m e f (x, y) = 2 x^{4} - x^{2} + 10 y^{2}

e x t re m e f (x) = x^{3} - 3 x^{2} - 9 x + 15

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

e x t re m e x^{\frac{1}{3}} (x + 4)