每日一题_191102

Posted 2020-12-01 math521

已知函数(f(x)=x{ln}x+dfrac{1}{2}ax^3-ax^2),(ainmathbb{R}).
((1)) 当(a=0)时,求(f(x))的单调区间;
((2)) 若函数(g(x)=dfrac{f(x)}{x})存在两个极值点(x_1,x_2),求(g(x_1)+g(x_2))的取值范围.
解析:
((1)) 当(a=0)时,对(f(x))求导可得[f'(x)=1+{ln}x,x>0.]
此时(f(x))在(left(0,dfrac{1}{mathrm{e}} ight))单调递减,在(left[dfrac{1}{mathrm{e}},+infty ight))单调递增.
((2)) 由题[ g(x)={ln}x+dfrac{1}{2}ax^2-ax,x>0.]对(g(x))求导可得[ g'(x)=dfrac{1}{x}+ax-a=dfrac{ax^2-ax+1}{x},x>0.]
若要使得(g'(x))在((0,+infty))有两个变号零点,需且仅需(g'left(dfrac{1}{2} ight)<0),即(a>4).根据韦达定理可得[ x_1+x_2=dfrac{1}{2},x_1x_2=dfrac{1}{a}.]
所以[ egin{split} &g(x_1)+g(x_2) =&{ln}(x_1x_2)+dfrac{1}{2}acdotleft(x_1^2+x_2^2 ight)-aleft(x_1+x_2 ight) =&{ln}(x_1x_2)+dfrac{1}{2}acdotleft[left(x_1+x_2 ight)^2-2x_1x_2 ight]-aleft(x_1+x_2 ight) =&-{ln}a+dfrac{1}{2}acdotleft(dfrac{1}{4}-dfrac{2}{a} ight)-dfrac{a}{2} =&-{ln}a-dfrac{3}{8}a-1. end{split}]
显然,上述表达式是关于(a)的单调递减函数,其中(a>4).因此所求表达式的取值范围为(left(-infty,-2{ln}2-dfrac{5}{2} ight)).