设y=log2 (x+1)上任意一点(x,y), 其关于x=1的对称点是(x‘,y’)有下面的关系：y' = y(x' +x)/2 = 1 ==> x= 2-x'y' = y=log2 (x+1)=log2 ((2-x')+1) =log2 (3-x')y=log2 (x+1)关于x=1对称的函数是 y = log2 (3-x)
y=log2 (-X-1)
