已知函数\(f(x)=\dfrac{x^2+2{\ln}x+3}{x}+m\),若\(\exists x_0\in\left[\dfrac{1}{4},+\infty \right)\),使得\(f(f(x_0))=x_0\),则\(m\)的取值范围是\(\underline{\qquad\qquad}\).
解析:
由于\(x_0\)满足\(f(f(x_0))=x_0\),所以\(x_0\)是\(f(x)\)的稳定点,又因为$$
f'(x)=\dfrac{x2-1-{\ln}x2}{x^2}\geqslant 0.$$因此\(x_0\)为\(f(x)\)的不动点,因此题意即$$\exists x_0\in \left[ \dfrac{1}{4},+\infty \right), f(x_0)=x_0.$$
所以问题转化为求函数$$
m(x)=x-\dfrac{x^2+2{\ln}x+3}{x},x\in\left[ \dfrac{1}{4},+\infty\right).$$的值域问题.易求得\(m\)的取值范围为\(\left[ -2\sqrt{\mathrm{e}},0\right)\).
