Consider the function f(x) satisfying the identity f(x)+fx−1x=1+x∀x∈R−{0,1}, and g(x)=2f(x)−x+1The domain of y=g(x)  is The range of y = g(x) is The number of roots of the equation g(x) = 1 is

f(x)+fx−1x=1+x            (1) In (1), replace x by x−1x. Then fx−1x+fx−1x−1x−1x=1+x−1xor   fx−1x+f11−x=1+x−1x                     (2)Now, from (1) - (2), we havef(x)−f11−x=x−x−1x               (3) In (3), replace x by 11−x. Then f11−x−fx−1x=11−x−11−x−111−xor   f11−x−fx−1x=11−x−x               (4)Now, from (1) + (3) + (4), we have2f(x)=1+x+x−x−1x+11−x−xor   f(x)=x3−x2−12x(x−1)g(x)=x3−x2−1x(x−1)−x+1=x2−x−1x(x−1)Now, for y=g(x), we must have x2−x−1x(x−1)≥0or   x−1−52x−1+52x(x−1)≥0or   x∈−∞,1−52∪(0,1)∪1+52,∞f(x)+fx−1x=1+x            (1) In (1), replace x by x−1x. Then fx−1x+fx−1x−1x−1x=1+x−1xor   fx−1x+f11−x=1+x−1x                     (2)Now, from (1) - (2), we havef(x)−f11−x=x−x−1x               (3) In (3), replace x by 11−x. Then f11−x−fx−1x=11−x−11−x−111−xor   f11−x−fx−1x=11−x−x               (4)Now, from (1) + (3) + (4), we have2f(x)=1+x+x−x−1x+11−x−xor   f(x)=x3−x2−12x(x−1)y=g(x)=x2−x−1x(x−1) or (y−1)x2+(1−y)x+1=0Now, x is real, Therefore, D≥0 or (1−y)2−4(y−1)≥0or  (y−1)(y−5)≥0or  y∈(−∞, 1] ∪ [5, ∞)f(x)+fx−1x=1+x            (1) In (1), replace x by x−1x. Then fx−1x+fx−1x−1x−1x=1+x−1xor   fx−1x+f11−x=1+x−1x                     (2)Now, from (1) - (2), we havef(x)−f11−x=x−x−1x               (3) In (3), replace x by 11−x. Then f11−x−fx−1x=11−x−11−x−111−xor   f11−x−fx−1x=11−x−x               (4)Now, from (1) + (3) + (4), we have2f(x)=1+x+x−x−1x+11−x−xor   f(x)=x3−x2−12x(x−1)g(x)=1, or x2−x−1x(x−1)=1 or −x−1=−x, which has no solution.