已知函数\(f(x)\)是定义域为\(R\)的奇函数，\(f(x+1)=f(-x+1)\)，且当\(0\leqslant x\leqslant 1\)时，\(f(x)=\tan x\)，则下列结论正确的是\((\quad)\)
A. \(f(-\dfrac{1}{2})< f(3)< f(\dfrac{3}{2})\)
B. \(f(-\dfrac{1}{2})< f(\dfrac{3}{2})< f(3)\) C. \(f(3)< f(\dfrac{3}{2})< f(-\dfrac{1}{2})\) D. \(f(3)< f(-\dfrac{1}{2})< f(\dfrac{3}{2})\)
