If (f(x))2×f1−x1+x=64x; ∀x∈Df, thenf (x) is equal toThe domain of f (x) isThe value of f (9/7) is

(f(x))2f1−x1+x=64x          (1) Putting 1−x1+x=y or x=1−y1+y, we getf1−y1+y2⋅f(y)=641−y1+yor   f(x)⋅f1−x1+x2=641−x1+x                 (2)Squaring (1) and dividing by (2), f(x)4f1−x1+x2f(x)f1−x1+x2=(64x)2641−x1+xor  {f(x)}3=64x21+x1−xor  f(x)=4x2/31+x1−x1/3 (f(x))2f1−x1+x=64x          (1) Putting 1−x1+x=y or x=1−y1+y, we getf1−y1+y2⋅f(y)=641−y1+yor   f(x)⋅f1−x1+x2=641−x1+x                 (2)Squaring (1) and dividing by (2), f(x)4f1−x1+x2f(x)f1−x1+x2=(64x)2641−x1+xor  {f(x)}3=64x21+x1−xor  f(x)=4x2/31+x1−x1/3 (f(x))2f1−x1+x=64x          (1) Putting 1−x1+x=y or x=1−y1+y, we getf1−y1+y2⋅f(y)=641−y1+yor   f(x)⋅f1−x1+x2=641−x1+x                 (2)Squaring (1) and dividing by (2), f(x)4f1−x1+x2f(x)f1−x1+x2=(64x)2641−x1+xor  {f(x)}3=64x21+x1−xor  f(x)=4x2/31+x1−x1/3or  x=f(9/7)=−4(9/7)2/3(2)