If the function f:[1,infinite)-->[1,infinite) is defined by f(x)=2^x(x-1), then f^-1 (x) is .

Answers (1)

Solution:   Let       y=2^x(x-1)Rightarrow log_2y=x(x-1)

                                 x^2-x-log_2y=0

                     Rightarrow       f^-1(y)=x=frac12(1pm sqrt1+4log_2y),

                    Rightarrow     f^-1(x)=frac12(1+sqrt1+4log_2x),  as x> 0

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