2.
定义在\(R\)上的函数\(f(x)\)对任意\(x_{1}\)、\(x_{2}(x_{1}\neq x_{2})\)都有\( \dfrac {f(x_{1})-f(x_{2})}{x_{1}-x_{2}} < 0\),且函数\(y=f(x-1)\)的图象关于\((1,0)\)成中心对称,若\(s\),\(t\)满足不等式\(f(s^{2}-2s)\leqslant -f(2t-t^{2})\),则当\(1\leqslant s\leqslant 4\)时,\( \dfrac {t-2s}{s+t}\)的取值范围是\((\) \()\)