Question

求解一道高中数学题··

设f(x)是周期函数，且最小正周期为2，且f(1+x)=f(1-x),当-1≤x≤0时，f（x)=--x.试求函数f(X)在区间【-1,3】的表达式

Answer 1

f(x)是周期函数，且最小正周期为2，
且f(1+x)=f(1-x),
∴f(x)=f(2+x)=f[1-(1+x)]=f(-x),
∴f(x)是偶函数。
0<=x<=1时-1<=-x<=0,
f(x)=f(-x)=-(-x)=x.
1<=x<=2时-1<=x-2<=0,
f(x)=f(x-2)=-(x-2)=2-x.
2<=x<=3时0<=x-2<=1,
f(x)=x-2.
综上，f(x)={-x,(-1<=x<=0);
{x,(0<=x<=1);
{2-x,(1<=x<=2);
{x-2,(2<=x<=3).