extreme f(x)=x^{(2/5)}(7/(2+x))

Lösung

extrempunkte

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

Minimum (0, 0), Maximum (\frac{4}{3}, \frac{7 \cdot 4 ^{\frac{2}{5}} \cdot 3 ^{\frac{3}{5}}}{10})

Schritte zur Lösung

f^{'} (x) = \frac{7 ( - 3 x + 4 )}{5 x ^{\frac{3}{5}} ( x + 2 ) ^{2}}

Intervalle finden:

Absteigend

: - \infty < x < - 2,

Absteigend

: - 2 < x < 0,

Ansteigend

: 0 < x < \frac{4}{3},

Absteigend

: \frac{4}{3} < x < \infty

Stecke

x = 0

x^{\frac{2}{5}} \frac{7}{2 + x} : 0

Minimum (0, 0)

Stecke

x = \frac{4}{3}

x^{\frac{2}{5}} \frac{7}{2 + x} : \frac{7 \cdot 4 ^{\frac{2}{5}} \cdot 3 ^{\frac{3}{5}}}{10}

Maximum (\frac{4}{3}, \frac{7 \cdot 4 ^{\frac{2}{5}} \cdot 3 ^{\frac{3}{5}}}{10})

Minimum (0, 0), Maximum (\frac{4}{3}, \frac{7 \cdot 4 ^{\frac{2}{5}} \cdot 3 ^{\frac{3}{5}}}{10})

e x t re m e Q (x, y) = 5 x + 3 x y

e x t re m e f (x, y) = x^{2} + y^{2} - 10 x + 26

e x t re m e f (x) = x^{3} - 9 x^{2} + 10

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

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