7. 已知函数f(x)={−2x+1,x<0−x2+2x+1,x⩾0f(x)=\left\{{\left.\begin{array}{l}{-2x+1,x<0}\\ {-{x^2}+2x+1,x\geqslant 0}\end{array}\right.}\right.f(x)={−2x+1,x<0−x2+2x+1,x⩾0,则f(x)f(x)f(x)的单调递增区间为 ___(0,1)(0,1)(0,1)___.