extreme f(x)=x^2-9ln(x)

Lösung

extrempunkte

f (x) = x^{2} - 9 ln (x)

Minimum (\frac{3}{2}, \frac{9}{2} - 9 (ln (3) - \frac{1}{2} ln (2)))

Schritte zur Lösung

f^{'} (x) = 2 x - \frac{9}{x}

Intervalle finden:

Absteigend

: 0 < x < \frac{3}{2},

Ansteigend

: \frac{3}{2} < x < \infty

Stecke

x = \frac{3}{2}

x^{2} - 9 ln (x) : \frac{9}{2} - 9 (ln (3) - \frac{1}{2} ln (2))

Minimum (\frac{3}{2}, \frac{9}{2} - 9 (ln (3) - \frac{1}{2} ln (2)))

e x t re m e f (x) = (\frac{1}{12} x^{3} + x^{2} + 3 x + \frac{26}{3})

e x t re m e f (x) = - 5 x^{3} + 45 x

e x t re m e f (x) = - 0.020724 x^{2} + 0.3692 x - 0.6287

e x t re m e f (x) = (x^{2} - 12) \cdot e^{- 2 x}

e x t re m e f (x) = 49 x^{2} - 10 x + 17