For the function f:[1, ) [1, ) defined by f(x)=(x-1)^4+1 , among the two statements: (I) The set S= x [1, ): f(x)=f⁻¹(x) contains exactly two elements, and (II) The set S= x [1, ): f(x)=f⁻¹(x+1) is an empty set,

Step-by-step solution

Given f(x) = (x-1)^4 + 1 for x 1 . To find the inverse function f⁻¹(x) , let y = (x-1)^4 + 1 . (x-1)^4 = y-1 x-1 = (y-1)^ 1/4 x = (y-1)^ 1/4 + 1 . Thus, f⁻¹(x) = (x-1)^ 1/4 + 1 . Evaluating the first statement, since f'(x) = 4(x-1)^3 0 for x 1 , f(x) is strictly increasing. For a strictly increasing function, the solutions to f(x) = f⁻¹(x) are the same as the solutions to f(x) = x . (x-1)^4 + 1 = x (x-1)^4 - (x-1) = 0 . Let t = x-1 0 . t^4 - t = 0 t(t-1)(t^2+t+1) = 0 . Since t 0 , the real roots are t = 0 and t = 1