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

Lösung

extrempunkte

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

Maximum (0, 0), Minimum (\frac{4}{15}, 3 (\frac{4}{15})^{\frac{5}{3}} - 2 (\frac{4}{15})^{\frac{2}{3}})

Schritte zur Lösung

f^{'} (x) = \frac{- 4 + 15 x}{3 x ^{\frac{1}{3}}}

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Absteigend

: 0 < x < \frac{4}{15},

Ansteigend

: \frac{4}{15} < x < \infty

Stecke

x = 0

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

Maximum (0, 0)

Stecke

x = \frac{4}{15}

3 x^{\frac{5}{3}} - 2 x^{\frac{2}{3}} : 3 (\frac{4}{15})^{\frac{5}{3}} - 2 (\frac{4}{15})^{\frac{2}{3}}

Minimum (\frac{4}{15}, 3 (\frac{4}{15})^{\frac{5}{3}} - 2 (\frac{4}{15})^{\frac{2}{3}})

Maximum (0, 0), Minimum (\frac{4}{15}, 3 (\frac{4}{15})^{\frac{5}{3}} - 2 (\frac{4}{15})^{\frac{2}{3}})

e x t re m e g (x) = - x^{3} + 3 x - 10

e x t re m e f (x) = \frac{x ^{4} - 5 x + 7}{x ^{2} + 1}

e x t re m e f (x) = 5 - 6 x^{2}

e x t re m e y = \frac{- 8 - 8 x ^{2}}{2 x ( 1 + x )}

e x t re m e f (t) = \frac{e ^{t}}{t ^{2} + 1}