extreme f(x)=x^2log_{9}(x)

解答

极值点

f (x) = x^{2} lo g_{9} (x)

极小值 \frac{1}{9 ^{\frac{1}{4 l n ( 3 )}}}, \frac{lo g _{9} ( \frac{1}{9 ^{\frac{1}{4 l n ( 3 )}}} )}{9 ^{\frac{1}{2 l n ( 3 )}}}

求解步骤

f^{'} (x) = 2 x lo g_{9} (x) + \frac{x}{2 ln ( 3 )}

Find intervals:

递减

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

递增

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

将

x = \frac{1}{9 ^{\frac{1}{4 l n ( 3 )}}}

代入

x^{2} lo g_{9} (x) : \frac{lo g _{9} ( \frac{1}{9 ^{\frac{1}{4 l n ( 3 )}}} )}{9 ^{\frac{1}{2 l n ( 3 )}}}

极小值 \frac{1}{9 ^{\frac{1}{4 l n ( 3 )}}}, \frac{lo g _{9} ( \frac{1}{9 ^{\frac{1}{4 l n ( 3 )}}} )}{9 ^{\frac{1}{2 l n ( 3 )}}}

e x t re m e f (x) = 2000 + 530 (ln (x))

e x t re m e f (x) = \frac{3 x + 4}{6 x + 1}

e x t re m e f (x) = 3 x^{4} - 96 x^{2} + 1

e x t re m e f (x) = 0.003 x^{2} + 4.2 x - 50

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