Yes, if p is a point of period 1 for f, then it is a fixed point for f and so for . Thus, if has only one fixed point, then so does f. The only thing you seem to have left undone is the number of points in which the graph of f intersects the graph of the identity function. But if p lies on both y= f(x) and y= x, then y= f(p)= p and p is a fixed point. Since f has only one fixed point, in how many pointscany= f(x) intersect y= x?