extreme f(x)=x^{1/2}-x^{3/2}

Lösung

extrempunkte

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

Minimum (0, 0), Maximum (\frac{1}{3}, (\frac{1}{3})^{\frac{1}{2}} - (\frac{1}{3})^{\frac{3}{2}})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: 0 < x < \frac{1}{3},

Absteigend

: \frac{1}{3} < x < \infty

Stecke

x = 0

x^{\frac{1}{2}} - x^{\frac{3}{2}} : 0

Minimum (0, 0)

Stecke

x = \frac{1}{3}

x^{\frac{1}{2}} - x^{\frac{3}{2}} : (\frac{1}{3})^{\frac{1}{2}} - (\frac{1}{3})^{\frac{3}{2}}

Maximum (\frac{1}{3}, (\frac{1}{3})^{\frac{1}{2}} - (\frac{1}{3})^{\frac{3}{2}})

Minimum (0, 0), Maximum (\frac{1}{3}, (\frac{1}{3})^{\frac{1}{2}} - (\frac{1}{3})^{\frac{3}{2}})

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

e x t re m e f (x) = \frac{y - 800}{- 2}

e x t re m e (x - 1) e^{\frac{1}{x}}

e x t re m e f (x) = x^{4} + x^{2} (y - 2) + 3 (y - 1)^{2}

e x t re m e f (x, y) = 2 e^{x} ln (1 + y)