已知函数\(f(x)=x{\ln}x+\dfrac{1}{2}ax^3-ax^2\),\(a\in\mathbb{R}\).
\((1)\) 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\(\left(0,\dfrac{1}{\mathrm{e}}\right)\)单调递减,在\(\left[\dfrac{1}{\mathrm{e}},+\infty \right)\)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\(g'\left(\dfrac{1}{2}\right)<0\),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\[
\begin{split}
&g(x_1)+g(x_2)\\
=&{\ln}(x_1x_2)+\dfrac{1}{2}a\cdot\left(x_1^2+x_2^2\right)-a\left(x_1+x_2\right)\\
=&{\ln}(x_1x_2)+\dfrac{1}{2}a\cdot\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-a\left(x_1+x_2\right)\\
=&-{\ln}a+\dfrac{1}{2}a\cdot\left(\dfrac{1}{4}-\dfrac{2}{a}\right)-\dfrac{a}{2}\\
=&-{\ln}a-\dfrac{3}{8}a-1.
\end{split}\]
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\(\left(-\infty,-2{\ln}2-\dfrac{5}{2}\right)\).
来源：https://www.cnblogs.com/Math521/p/11785963.html